Co-ordinate Geometry

P on x-axis, since ordinate is zero.

Q on y-axis, since abscissa is zero.

R on x-axis, since ordinate is zero.

S on y-axis, since abscissa is zero.

(i) Since each side of square is 2a.

Coordinates of A are (0, 0), since it coincides with origin.

Coordinates of B are (2a, 0), for a point along x-axis ordinate is zero.

Coordinates of C are (2a, 2a), since this point is equi-distance from x-axis and y-axis.

Coordinates of D are (0, 2a), since abscissa is zero and ordinate is 2a.

(ii) Each side of square is a units.

Coordinates of A are (a, a), since this point lies in Ist coordinate.

Coordinates of B are (-a, a), since this point lies in IInd coordinate.

Coordinates of C are (-a, -a), since this point lies in IIIrd coordinate.

Coordinates of D are (a, -a), since this point lies in IVth coordinate.

Since PQ is the base of two equilateral triangles with side 2a and mid-point of PQ is at origin.

Therefore point R lies on positive x-axis and point R’ lies on negative y-axis.

OR² = (2a)²- a²

OR² = 4a²- a²

OR = √3a

Therefore coordinates of R are ( √3a, 0) and R’ (0, √3a)

(i) (-6,7) and (-1, -5)

Distance =

Distance =

Distance =

Distance =

Distance == 13 units

Thus, the distance between the points (-6,7) and (-1, -5) is 13 units

(ii) (a + b, b + c) and (a - b, c - b)

Distance =

Distance =

Distance =

Distance =

Distance == units

Thus, the distance between these points is (a + b, b + c) and (a - b, c - b) is units.

(iii) (a sin a, - b cos a) and (-a cos a, b sin a)

Distance =

(iv) (a, 0) and (0, b)

Distance =

Distance =

Distance =
Thus, the distance between points (a, 0) and (0, b) is

Distance =

= 10

a² - 2a - 8 = 0

a² - 4a + 2a – 8 = 0

a(a - 4) + 2(a - 4) = 0

(a - 4) (a + 2) = 0

a = 4 and a = -2

Distance from point (2, 1) = Distance from point (1, -2)

Square roots are cancelled, therefore

Given: the distances of the point (x, y) from (-3, 0) as well as from (3, 0) are 4.

To find: the values of x, y

Solution:

distances of the point (x, y) from (-3, 0) is

Distance =

…………….. (1)


distances of the point (x, y) from (3, 0) is

Distance =

…………….. (2)


Subtract eq 1 from eq 2 to get,

⇒ -6x + x² + y² - (6x + x² + y² ) = 7 - 7

⇒ -6x + x² + y² - 6x - x² - y² = 0

⇒ -12 x = 0

⇒ x = 0

Putting the value of x in eq 1 we get,

6x + x² + y² = 7

⇒ 6(0) + 0² + y² = 7

⇒ y² = 7

⇒ y=

Hence, X=0, y=

Let the ordinate of other end is k

Distance =

On squaring both sides, we get

; ;

Therefore ordinates are 3, -9

Given: the points A(- 4, -1), B(-2, - 4), C(4, 0) and D(2, 3)

To prove: the points are the vertices points of a rectangle.

Solution:


Vertices of rectangle ABCD are: A(- 4, -1), B(-2, - 4), C(4, 0) and D(2, 3)

Length of sides=

Length of side AB = = = units

Length of side BC = = = = units

Length of side CD = = = units

Length of side AD = = = = units

Length of diagonal BD = = = units

Length of diagonal AC = = = units

Since opposite sides are equal and diagonal are equal. Therefore given vertices are the vertices of a rectangle.

Vertices of a parallelogram ABCD are: A (1,- 2), B (3, 6), C (5, 10) and D (3, 2) Length of side AB =

Length of side AB = = √(4+64)= √68 units

Length of side BC = = √(4+16) = √20 units

Length of side CD = = √(4+64)= √68 units

Length of side DA = = √(4+16) = √20 units

Length of diagonal BD = = √16= 4 units

Length of diagonal AC = = √(16+144) = √160 units

Opposite sides of the quadrilateral formed by the given four points are equal i.e. (AB = CD) & (DA = BC)
Also, the diagonals BD & AC are unequal.
Therefore, the given points form a parallelogram.

Vertices of a square ABCD are: A (1, 7), B (4, 2), C (-1, -1) and D (-4, 4) Length of side AB =

Length of side AB = = = units

Length of side BC = = = units

Length of side CD = = = units

Length of side DA = = = units

Length of diagonal BD = = = units

Length of diagonal AC = = = units

Since opposite sides are equal and diagonal are equal. Therefore given vertices are the vertices of a square.

Vertices of a triangle ABC are: A(3, 0), B(6, 4) and C (- 1, 3)

Length of side AB =

Length of side AB = = = units

Length of side BC = = = units

Length of side AC = = = units

Since AB = AC, therefore triangle is an isosceles.

BC² = AB² + AC²

(√50)² = (√25)² + (√25)²

50 = 25 + 25

50 = 50

Since BC² = AB² + AC²; therefore given triangle is right angled triangle.

Solution:

Vertices of a triangle ABC are: A(2, -2), B(-2, 1) and C(5, 2)

Length of side AB =

Length of side AB = = = units

Length of side BC = = = units

Length of side AC = = = units

Since AB = AC, therefore triangle is an isosceles.

BC² = AB² + AC²

(√50)² = (√25)² + (√25)²

50 = 25 + 25

50 = 50

Since BC² = AB² + AC²; therefore given triangle is right angled triangle.

Area of right angled triangle =

Area of right angled triangle = square units

Length of hypotenuse (BC) = = units

Vertices of a triangle ABC are: A(2 a, 4 a), B(2 a, 6 a) and C

Length of side AB =

Length of side AB = = = 2a units

Length of side BC = = a + a units

Length of side AC = = a + a units

The given vertices are not the vertices of an equilateral triangle

Let the Vertices of a triangle ABC are: A(2, 3), B(-4, -6) and C(1, 3/2)

Length of side AB =

Length of side AB = = = units

Length of side BC = = = units

Length of side AC = = = units

The given vertices do not form a triangle, since sum of two sides of a triangle are not greater than third side.

Given: An equilateral triangle has two vertices at the points (3, 4) and (-2, 3)
To find: the coordinates of the third vertex.
Solution: Let the Vertices of a triangle ABC are A(3, 4) and B (-2, 3), and C(x, y),
Since it is equilateral triangle,
AB=AC=BC
Where AB,AC and BC are lengths of sides of the given triangle.
To find the length of a side use distance formula ,

Length of side AB =

Length of side AB = = = units

Length of side BC = units

Length of side AC = units
Now AB=AC

⇒(AB)² = (BC)²

=

(AB)² = (AC)²

=

On subtracting eqn (2) from (1), we get

...... (3)

(AC)² = (BC)²

..... (4)

Solving equations (3) and (4), we get

; y =

Therefore coordinates of C are (, )

Let the Vertices of a quadrilateral are: A(2, -1), B(3, 4), C(-2, 3) and D(-3, -2)

Length of side AB =

Length of side AB = = = units

Length of side BC = = = units

Length of side CD = = = units

Length of side DA = = = units

Since all sides are of equal length, therefore it is a rhombus.

Vertices of an isosceles are: A(2, 0) and B(2, 5).

Let the third vertex is P(x, y)

Using distance formula =

Length of side PA = units

Length of side PB = units

Since PA = PB

On squaring both sides, we get

,

Also, PA = 3

On squaring both sides, we get

On substituting ,

Using quadratic formula:

Therefore coordinates of third vertex are: ;

Since the point is on x-axis, therefore coordinate of y-axis is zero.

Therefore the point is P(k, 0) which is equidistance from A(5, 9) and B(- 4, 6)

PA = PB

On squaring both sides

Therefore coordinate is (27, 0)

Vertices are: A(- 2, 5), B(0, 1) and C(2, - 3)

Using distance formula =

Length of side AB = = = = units

Length of side BC = = = = units

Length of side AC = = = = units

Since length of AB + BC = AC, therefore points are collinear.

Let the coordinates of Point Q are (x, y) and coordinates of origin O are (0, 0)

Since OP = OQ

Points are: P (-3, 2), Q(x, y) and O (0, 0)

Q is the mid point

Therefore coordinates are (3, -2)

Since the point is on y-axis, therefore coordinate of x-axis is zero.

Therefore the point is P(0, k) which is equidistance from A(2, 3) and B(- 4, 1)

PA = PB

On squaring both sides, we get

Therefore coordinate is (0, -1)

Consider A(3, 4), B (3, 8) and C(9, 8).

Let the coordinates of fourth vertex are D (x, y)

In a parallelogram diagonals bisect each other

Coordinate of mid point of AC =

Therefore coordinates of mid point of AC are (6, 6)

Coordinate of mid point of BD =

Coordinates of point D are

Therefore coordinates of fourth vertex D are (9, 4)

Vertices of triangle are A(-2, -3), B(- 1, 0), C(7, - 6)

Let the coordinates of P are (x, y)

PA = PB =PC

PA = PB

On squaring both sides, we get

…………(1)

PA = PC

On squaring both sides, we get

…………(2)

On solving equations (1) & (2), We get

X = 3 and y = -3

Therefore coordinates are (3, -3)

Since the abscissa of first coordinate is zero, therefore this point lies on y-axis. Ordinate of second point is zero, therefore this point lies on y-axis. We know that both the axes are perpendicular to eachother, therefore the angle between these points is 90°.

Coordinates of points on a circle are A(2,1), B(5,-8) and C(2,-9).

Let the coordinates of the centre of the circle be O(x, y)

Using distance formula =

Since the distance of the points A, B and C will be equal from the center, therefore

⇒ OA = OB

On squaring both sides, we get

⇒ x² + 4 - 4x + y² + 1 - 2y = x² + 25 - 10x + y² + 64 + 16y

⇒ 6x - 18y - 84 = 0

⇒ x - 3y - 14 = 0 ------------- (1)

Similarly, OC = OB

⇒ x² + 4 - 4x + y² + 81 + 18y = x² + 25 - 10x + y² + 64 + 16y

⇒ 6x - 2y - 4 = 0

⇒ 3x - y - 2 = 0 ------------- (2)

By solving equations (1) and (2), we get x = -1, y = -5

So, the coordinates of the centre of the circle is (-1, -5).

Radius of the circle = OA =

=

=

= units

Let the point is P(0, 2) which is equidistance from A(3, k) and B(k, 5)

PA = PB

On squaring both sides, we get

Let ABCD is a square with A(5, 4) and C(1, -6).

Let the coordinates of B are (x, y)

Using distance formula =

⇒ AB = BC

On squaring both sides, we get

……………………….(1)

In ΔABC, Using Pythagoras theorem

AC² = AB² + BC²

AC² = 2BC² [Since AB = BC]

On squaring both sides, we get

………………..(2)

On substituting value of from equation (1) in equation (2), we get

On solving we get y = -3, -1

Substituting these values of y in eqn 1, we get x = 8, -2

Therefore other coordinates are B(8, -3). And D(-2,1)

Vertices of the rhombus are: A(-3, 2), B(-5, -5), C(2, -3) and D(4, 4)

We know that diagonals of a rhombus bisect each other, therefore point of intersection of diagonals is:

Abscissa of Mid point of AC =

Ordinate of Mid point of AC =

Abscissa of Mid point of BD =

Ordinate of Mid point of BD =

Since the diagonals AC and BD bisect each other at O, therefore it is a rhombus.

Length of diagonal AC = = = = units

Length of diagonal BD = = = = units

Area of rhombus = = = 45 sq units

Area of rhombus is 45 sq units

Coordinates of points on a circle are A (3, 0), B(-1, -6) and C(4,-1)

Let the coordinates of the centre of the circle be O(x, y)

Using distance formula =

Since the distance of the points A, B and C will be equal from the center, therefore

⇒ OA = OC

On squaring both sides, we get

=

⇒ x² + 9 - 6x + y² = x² + 2x +1 + y² + 36 + 12y

⇒ -8x - 12y = 28

⇒ 2x + 3y = -7------------- (1)

Similarly, OC = OB

On squaring both sides, we get

=

⇒ x² + 16 - 8x + y² + 1 + 2y = x² + 1 +2x + y² + 36 + 12y

⇒ -10x - 10y = 20

⇒ x + y = -2 ------------- (2)

Solving eqn (1) and (2), we get

x = 1; y = -3

Coordinates of circum center are (1, -3)

Circum radius of the circle = OA =

=

= units

points A(7, 6) and B(-3, 4) are equidistance from point P.

Let the coordinates of point are P(x, 0)

Using distance formula =

⇒ PA = PB

On squaring both sides, we get

=

Therefore coordinates are (3, 0)

Vertices of a quadrilateral are A(5,6), B(1,5), C(2,1) and D(6,2).

Using distance formula =

AB = = = units

BC = = = units

CD = = = units

DA = = = units

AB = BC = CD = DA

BD = = = units

AC = = = units

All the four sides of the quadrilateral are equal and diagonals are of equal length. Therefore, the given vertices form a square.

The Vertices of a quadrilateral are A(5,6), B(1,5), C(2,1) and D(6,2).

Using distance formula =

AB = = = units

BC = = = units

CD = = = units

DA = = = units

AB = BC = CD = DA

BD = = = units

AC = = = units

All the four sides of the quadrilateral are equal and diagonals are of equal length. Therefore, the given vertices form a square.

Points A(-2, 5) and B(2, -3) are equidistant from point P.

Let the coordinates of point are P(x, 0)

Using distance formula =

⇒ PA = PB

On squaring both sides, we get

=

Hence, coordinates are (-2, 0).

Coordinates are P(6,-1), Q(1, 3) and R(x, 8)

Using distance formula =

⇒ PQ = QR

On squaring both sides, we get

=

=

On solving above equation, we get

Therefore x = -3, 5

Vertices of a quadrilateral are A (0, 0), B(5, 5) and C(-5, 5)

Using distance formula =

AB = = = units

BC = = = units

CA = = = units

Since AB = CA

Using Pythagoras theorem

BC² = AC² + AB²

100 = 50 + 50

100 = 100

Therefore vertices are of right isosceles triangle.

Coordinates are P(x, y), A(5, 1) and B (1, 5)

Using distance formula =

⇒ PA = PB

On squaring both sides, we get

=

=

=

proved

Coordinates are Q(0, 1), P (5, -3) and R (x, 6),

Using distance formula =

⇒ QP = QR

On squaring both sides, we get

=

=

=

proved

QR = = = units

PR = = = units

Given: the distance between the points P (2, -3) and Q (10, y) is 10 units.
To find: The value of y.
Solution:
Coordinates are P (2, -3) and Q (10, y)

We use distance formula to find the distance between two points.

Since PQ = 10 units
S0,

= 10

On squaring both sides, we get

= 100

⇒ = 100
⇒ 73 + y² + 6y = 100
⇒ 73 + y² + 6y -100 = 0
⇒ = 0
In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ y² + 9y - 3y - 27 = 0

⇒y(y + 9) - 3(y + 9) = 0

⇒(y - 3)(y + 9) = 0

⇒y = 3, -9

Coordinates of points on a circle are A(6, -6), B(3, -7) and C(3, 3).

Let the coordinates of the centre of the circle be O(x, y)

Using distance formula =

Since the distance of the points A, B and C will be equal from the center, therefore

⇒ OA = OC

On squaring both sides, we get

=

=

= ………………….(1)

Similarly, OA = OB

On squaring both sides, we get

=

=

= …………..(1)

Solving eqn (1) and (2), we get

x = 3; y = -2

Coordinates of circum center are (3, -2)

The coordinates are A(-1, 2) and C(3, 2).

Let the coordinates of the vertex B are (x, y)

AB = BC

Using distance formula =

On squaring both sides, we get

=

=

=

In ΔABC

AB² + BC² = AC² [Using Pythagoras theorem]

2AB² = AC² [Since AB = BC]

=

= 16

= 8

= 3

On substituting x = 1

= 3

= 0

= 0

= 0, 4

Other coordinates are (1, 0) and (1, 4)

(i) A (-1, - 2), B (1, 0), C (-1, 2), D (-3, 0)

Using distance formula:

AB = BC = CD = DA

All the four sides of the quadrilateral are equal and diagonals are of equal length. Therefore, the given vertices form a square.

(ii) A (-3, 5), B (3, 1), C (0, 3), D (-1, - 4)

Using distance formula:

Since all sides are of different length, therefore it is not a particular type of quadrilateral.

(iii) A (4, 5), B (7, 6), C (4, 3), D (1, 2)

Using distance formula

Coordinates of midpoint of diagonal AC
X=
Y =

Therefore coordinates of midpoint of AC are (4, 4)

Coordinates of midpoint of diagonal BD=
X = , Y =

Therefore coordinates of midpoint of AC are (4, 4)

Since diagonals bisect each other at same point therefore quadrilateral is a parallelogram.

The points are A(7, 1) and B(3, 5).

Coordinates of midpoint of line AB = X = , Y =

Therefore coordinates of midpoint of AB are (5, 3)

Slope of the line = = = = 2

Negative reciprocal of slope =

Equation of line Y= mX+C

Since diagonals bisect each other at same point therefore quadrilateral is a parallelogram.

Let the Vertices of a quadrilateral are: A(3, 0), B(4, 5), C(-1, 4) and D(-2, -1),

Length of side AB =

Length of side AB = = = units

Length of side BC = = = units

Length of side CD = = = units

Length of side DA = = = units

Length of diagonal AC = = = units

Length of side BD = = = units

Since all sides are of equal length, therefore it is a rhombus.

Area of Rhombus = = = = 24 sq. units

Points are A (3, 1), B (6, 4) and C (8, 6)

For sitting in a line three points must be collinear i.e AB + BC = AC

Length of side AB =

Length of side AB = = = = units

Length of side BC = = = = units

Length of side AC = = = units

AB + BC = AC

+ =

The points are collinear.

Points A(5, -2) and B(-3, 2) are equidistance from point P.

Let the coordinates of point are P(0, y)

Using distance formula =

⇒ PA = PB

On squaring both sides, we get

(0-5)² + (y+2)² = (0+3)² + (y-2)²

⇒ (-5)² + (y+2)² = (3)² + (y-2)²

⇒ 25 + y² + 4 + 4y = 9 + y² + 4 – 4y

⇒ 4y + 4y = 9 – 25

⇒ 8y = -16

⇒ y = -2

Therefore coordinates are (0, -2).

Coordinates of the points are A(3, 6) and B(-3, 4)

Let the point P(x, y) is equidistant from A and B

Using distance formula =

⇒ PA = PB

On squaring both sides, we get

=

Using distance formula =

⇒ AB = AC

On squaring both sides, we get

=

Vertices of a quadrilateral are A (7, 10), B(-2, 5) and C(3, -4)

Using distance formula =

AB = = = units

BC = = = units

AC = = = units

Since AB = BC

Using Pythagoras theorem

AC² = AB² + BC²

212 = 106 + 106

212 = 212

Therefore vertices are of right isosceles triangle.

Coordinates are A (7,-1) and B (6, 8)

The point P (x, 3) is equidistant.

Using distance formula =

⇒ PA = PB

On squaring both sides, we get

=

=

=

AP = = = units

Coordinates are P(8, -3) and Q(7,6)

The point A (3, y) is equidistant.

Using distance formula =

⇒ PA = QA

On squaring both sides, we get

=

=

= 1

AQ = = = units

Coordinates are A(0, -3) and B(0,3) and C(x, y)

Using distance formula =

⇒ AB = AC

On squaring both sides, we get

=

=

(1)

⇒ AB = BC

On squaring both sides, we get

=

=

(2)

On subtracting equation (2) from (1) we get

On substituting y = 0 in equation (1), we get

Therefore coordinates of third vertex are (), (),

Coordinates of points are P(2, 2) A(-2, k) and B(-2k, -3)

Using distance formula =

⇒ PA = PB

On squaring both sides, we get

=

=

0

0

0

0

-1, -3

AP = = = 5 units

Coordinates of points are A(0, 2), B(3, p) and C(p, 5)

Using distance formula =

⇒ AB = AC

On squaring both sides, we get

=

=

1

AB = = = units

Coordinates of points are A(3, k), B(k, 5) and P(k-1, 2)

Using distance formula =

⇒ PA = PB

On squaring both sides, we get

=

=

= 1 + 9

= 0

= 0

= 0

= 0

= 1, 5

Let our points be A(-1, 3) and B(4, -7) and required point be C( x, y)

Given that point divides internally in ratio of 3:4.

By section formula,

x = , y =

Here, m = 3 and n = 4

∴ x = , y =

∴ x= , y =

∴ x = , y =

Hence, the required point is C( , )

(i) (5, -6) and (- 7, 5),

Let our given points be A(5,-6) and B(-7, 5) and required points be C (x₁ , y₁ ) and D(x₂ , y₂)

The points of trisection of a line are points which divide into the ratio 1:2

By section formula,

x = , y =

For point C(x₁ , y₁ )

x₁ = , y₁= …Here m = 1 and n = 2

∴ x₁ = , y₁ =

∴ C (x₁ , y₁ ) ≡ (1, )

For point D(x₂ , y₂ )

X₂ = , y₂= …Here m = 2 and n = 1

∴ x₂ = , y₂ =

∴ D (x₂ , y₂)≡ (-3, )

Hence, the points of trisection of line joining given points are (1, ) and (-3, )

(ii) (3, -2) and (-3, -4)

Let our given points be A(3,-2) and B(-3, -4) and required points be C (x₁ , y₁ ) and D(x₂ , y₂)

The points of trisection of a line are points which divide into the ratio 1:2

By section formula,

x = , y =

For point C(x₁ , y₁ )

x₁ = , y₁= …Here m = 1 and n = 2

∴ x₁ = , y₁ =

∴ C (x₁ , y₁ ) ≡ ( , )

For point D(x₂ , y₂ )

X₂ = , y₂= …Here m = 2 and n = 1

∴ x₂ = , y₂ =

∴ D (x₂ , y₂)≡ (-1, )

Hence, the points of trisection of line joining given points are ( , ) and (-1, )

(iii) (2, -2) and (-7, 4)

Let our given points be A(2,-2) and B(-7, 4) and required points be C (x₁ , y₁ ) and D(x₂ , y₂)

The points of trisection of a line are points which divide into the ratio 1:2

By section formula,

x = , y =

For point C(x₁ , y₁ )

x₁ = , y₁= …Here m = 1 and n = 2

∴ x₁ = , y₁ =

∴ C (x₁ , y₁ ) ≡ ( , )

For point D(x₂ , y₂ )

X₂ = , y₂= …Here m = 2 and n = 1

∴ x₂ = , y₂ =

∴ D (x₂ , y₂)≡ (-4, 2)

Let our points of parallelogram be A(-2, -1), B(1, 0), C(4, 3) and D(1, 2) and mid point of diagonals be E(x,y)

We know that diagonals of parallelogram bisect each other.

Hence, we find mid point of AC.

By midpoint formula,

x = , y =

For point E(x, y)

x₁ = , y₁=

∴ x₁ = , y₁ =

∴ E (x, y) ≡ ( , 1 )

We know if the quadrilateral is parallelogram if opposite sides are equal.

Let our points be A(3, -2), B(4, 0), C(6, -3) and D(5, -5).

By distance formula,

XY =

For AB,

AB =

=

= units

For BC,

BC =

=

= units

For CD,

CD =

=

= units

For AD,

AD =

=

= units

Here, we observe that AB = CD and AD = BC, which means that the quadrilateral formed by lines joining by points, is parallelogram.

Let three vertices be A(-2, -1), B(1, 0) and C(4, 3) and fourth vertex be D(x, y)

It is given that quadrilateral joining these four vertices is parallelogram.

∴□ABCD is parallelogram

We know that diagonals of parallelogram bisect each other, ie midpoint of the diagonals coincide.

Let E(x_m , y_m) be the midpoint of diagonals AC and BD.

By midpoint formula,

x = , y =

For diagonal AC,

x_m = , y_m =

∴ x_m = , y_m =

∴ E(x_m , y_m) ≡ (1, 1)

For diagonal BD,

1 = , 1 =

∴ x = 2 – 1 , y = 2 – 0

∴ x = 1 and y = 2

Hence, our fourth vertex is D(1, 2)

Let three vertices be A(3, -4), B(-1, -3) and C(-6, 2) and fourth vertex be D(x, y)

It is given that quadrilateral joining these four vertices is parallelogram.

∴□ABCD is parallelogram

We know that diagonals of parallelogram bisect each other, ie midpoint of the diagonals coincide.

Let E(x_m , y_m) be the midpoint of diagonals AC and BD.

By midpoint formula,

x = , y =

For diagonal AC,

x_m = , y_m =

∴ x_m = , y_m =

∴ E(x_m , y_m) ≡ (, -1)

For diagonal BD,

= , -1 =

∴ x = -3 + 1 , y = -2 + 3

∴ x = -2 and y = 1

Hence, our fourth vertex is D(-2, 1)

Here, given points are A (-2, 2) and B ( 3, 7) and let the point dividing the line joining two points be C(2,y).

Let the ratio be m:n

By section formula,

x = , y =

For point C(2,y),

2 = …(1)

And y = …(2)

Solving 1 for finding ratio between m and n,

2 =

2(m + n) = 3m -2n

2m + 2n = 3m – 2n

∴ m = 4n

∴ =

∴ m : n = 4 : 1

Now solving for equation 2, where m = 4 and n = 1

y =

y =

∴ y =

∴ y =

∴ y =6

Hence, our point is (2, 6)

Here given vertices of triangle are A (-1, 3), B (1, -1) and C (5, 1).

Let D, E and F be the midpoints of the sides BC, CA and AB respectively.

We need to find length of median passing through A, ie distance between AD.

Let point D ≡ (x, y)

By midpoint formula,

x = , y =

For midpoint D of side BC,

x = , y =

∴ x = , y =

∴D(x , y) ≡ (3, 0 )

Now, by distance formula,

XY =

For AD,

AD =

∴ AD =

AD =

∴AD = 5 units

Hence, the length of the median through A is 5 units

Let A(x₁ , y₁), B(x₂ , y₂) and C(x₃ , y₃) be the vertices of triangle.

Let D(1, 1), E(2, -3) and F(3, 4) be the midpoints of sides BC, CA and AB respectively.

By midpoint formula.

x = , y =

For midpoint D(1, 1) of side BC,

1 = , 1 =

∴ = 2 and = 2 …(1)

For midpoint E(2, -3) of side CA,

2 = , -3 =

∴ = 4 and = -6 …(2)

For midpoint F(3, 4) of side AB,

3 = , 4 =

∴ = 6 and = 8 …(3)

Adding 1,2 and 3, we get,

= 2 + 4 + 6

And = 2 – 6 +8

∴ 2(= 12 and 2() = 2

∴ = 6 and = 1

+ 2 = 6 and + 2 = 2 …from 1

∴ = 4 and = 0

Substituting above values in 3,

4 + and 0 + = 8

∴ = 2 and = 89

Similarly for equation 2,

4 + = 6 and 0 + = -6

∴ = 2 and = -6

Hence the vertices of triangle are A(4 , 0), B(2 ,8) and C(0 ,-6)

Let in ∆ABC, A(1,1), B(x₁, y₁) and C(x₂, y₂).

Let D(-2, 3) and E(5, 2) be the midpoints of sides AB and AC respectively.

By midpoint formula.

x = , y =

For mid point D (-2, 3) of side AB,

-2 = , 3 =

1 + x₁ = -4 and 1 + y₁ = 6

x₁ = -5 and y₁ = 5

∴ B ( x₁, y₁) ≡ (-5, 5)

For midpoint E(5, 2) of side AC,

5 = , 2 =

1 + x₂ = 10 and = 4

∴ x₂ = 9 and = 3

∴ C(x₂, y₂) ≡ (9, 3)

Hence other two vertices are B (-5, 5) and C(9, 3)

Here y axis divides our line joined by the points (say) A (-2, -3) and B (3, 7).

Let coordinate of the point be C(0, y).

Here our x- coordinate is zero, as point C lie on x-axis.

Let y axis divide AB in ratio of m:n.

By section formula,

x = , y =

For point C(0, y) on line joined by the points A and B,

0 = …(1)

And, y = …(2)

Solving 1,

0(3m – 2n

∴ 3m = 2n

∴m:n = 2:3

Now solving for 2, for values m = 2 and n = 3,

y =

y =

∴ y = = 1

∴ C(0, y) ≡ (0, 1)

Let given points be A (-3, -1) and B (-8, -9).

Let the point C(-5, -21 /5) divide AB in ratio m:n.

By section formula,

x = , y =

For point C(-5, -21 /5) on the line joined by the points A and B.

-5 = …(1)

And, - = …(2)

Solving 1,

-5(m + n) = -8m - 3n

∴5m + 5n = 8m + 3n

∴ 2n = 3m

∴ =

Hence, ratio is 2:3.

Let A(3, 4) and B(k, 7) and midpoint be C(x, y) which lies on the line 2x+2y+1 = 0

By midpoint formula,

x = , y =

For point C(x, y),

x = , y = …(1)

Here, y = ,

Hence, substituting value of y in given equation of line,

2x + 2 × + 1 = 0

∴ 2x = -12

∴ x = -6

Now substituting value of x in equation(1), we get.

x =

-6 =

∴ -12 = 3 + k

∴ k = -15

Hence, the value of k is -15.

Let point be A(3, -1) and B(8, 9).

Let the line divide the line joining the points A and B in the ratio m:n at any point C(x, y)

By section formula,

x = , y =

For point C(x, y),

x = , y =

∴ x = , y =

Now, substituting value of x and y in equation x - y - 2 = 0,

- -2 = 0

= 0

∴ -3m +2n =0

∴ =

∴ m:n = 2:3

Hence, the line divides the line segment joining A and B in the ratio 2:3 internally.

(i) x-axis

Let our points be A(-2, -3) and B(5, 6).

Let point C(x, 0) divide the line formed by joining by the points A and B in ratio of m:n.

By section formula,

x = , y =

For point C(x, 0)

x = , 0 =

Solving for y coordinate,

0 =

∴ 6m -3n = 0

∴ 2m = n

∴ =

∴ m : n = 1 : 2

Now solving for x coordinate, with m = 1 and n = 2,

x =

∴ x=

∴ x =

Hence, the coordinates of required point is C( , 0)

(ii) y-axis.

Let our points be A(-2, -3) and B(5, 6).

Let point C(0, y) divide the line formed by joining by the points A and B in ratio of m:n.

By section formula,

x = , y =

For point C(0, y)

0 = , y =

Solving for x coordinate,

0 =

∴ 5m – 2n = 0

∴ =

∴ m : n = 2 : 5

Now solving for y coordinate, with m = 2 and n = 5,

y =

y =

∴ y =

Hence, the coordinates of required point is C( , )

Let given points be A(4, 5), B(7, 6), C(6, 3), D(3, 2) and let the intersection of diagonals be E(x_m , y_m )

By midpoint formula.

x = , y =

For midpoint of diagonal AC,

X₁ = , y₁ =

∴x₁ = = 5 , y_{1 =} = 4

∴ midpoint of diagonal AC is (x₁, y₁ ) ≡ (5, 4) …(1)

For midpoint of diagonal BD,

X₂ = , y₂ =

∴x₂ = = 5 , y ₂₌ = 4

∴ midpoint of diagonal BD is (x₂, y₂ ) ≡ (5, 4) …(2)

Here, from 1 and 2 we say that midpoint of both the diagonals intersect at same point, ie (5, 4)

But our intersection of diagonals is at E, which means that midpoint of diagonals intersect at single point, ie E(5, 4)

We know that if midpoints of diagonals intersect at single point, then quadrilateral formed by joining the points is parallelogram.

Hence, our □ABCD is parallelogram.

Now, we shall check whether □ABCD is rectangle.

If the lengths of diagonals are same, then given quadrilateral is rectangle.

By distance formula,

XY =

For diagonal AC,

AC =

=

= units

For diagonal BD,

AC =

=

= units.

Here, AC ≠ BD, hence □ABCD is not rectangle.

Let given points be A(4, 3), B(6, 4), C(5, 6) and D(3, 5).

By distance formula,

XY =

For AB,

AB =

=

= units.

For BC,

BC =

=

= units.

For CD,

CD =

=

= units.

For AD,

AD =

=

= units.

Here, we can observe that □ABCD is a parallelogram.

Now,

For diagonal AC,

AC =

=

= units.

For diagonal BD,

BD =

=

= units.

∴ AC = BD, which means diagonals are equal.

We know that quadrilateral in which all sides are equal and diagonals are equal, is a square.

∴□ABCD is a square.

Solution : Let the given points be A(-4, -1), B(-2, -4), C(4, 0) and D(2, 3).

Use distance formula,

For AB,

AB =

=

=

= units

For BC,

BC =

=

= units

For CD,

=

= units

For AD,

AD =

=

= units

Also, for diagonal AC,

AC =

= units

For diagonal BD,

BD =

=

= units

We can observe that AB = CD and BC = AD and also diagonal AC = BD.

We know that a quadrilateral whose opposite sides are equal and the diagonal are equal is rectangle.

∴ ABCD is a rectangle.

Here given vertices are A(-1,3), B (1,-1) and C(5,1) and let midpoints of BC, CA and AB be D,E and F respectively.

By midpoint formula.

x = , y =

For midpoint D of side BC,

x = , y =

x = , y =

∴midpoint of side BC is D(3, 0)

For midpoint E of side AB,

x = , y =

x = , y =

∴midpoint of side AB is E(2, 2)

For midpoint F of side CA,

x = , y =

x = , y =

∴midpoint of side CA is F(0, 1)

By distance formula,

XY =

For median AD,

AD =

=

=

= 5 units

For median BE,

BE =

=

= units.

For median CF,

CF =

=

= 5 units

Let A(a + b, a - b), B(2a + b, 2a - b), C(a - b, a + b) and fourth vertex be D(x, y).

It is given that □ABCD is parallelogram.

We know that diagonals of parallelogram bisect each other.

Let intersection of diagonals be E(x_m, y_m )

By midpoint formula.

x_m = , y_m =

For midpoint E of diagonal AC,

x_m = , y_m =

∴ x_m = a , y_m = a

∴E(x_m, y_m ) ≡ (a, a)

For diagonal BD,

a = , a=

∴ 2a = 2a + b +x , 2a = 2a – b +y

∴ x = -b and y = b

Hence, the fourth vertex is D(-b, b)

Let the vertices be A(3, 2), B(-1, 0), C(x₁, y₁ ) and D(x₂, y₂ ).

Let diagonals cut at E(2, - 5).

We know that mid points of diagonals of parallelogram coincide.

By midpoint formula.

x_m = , y_m =

For midpoint E of diagonal AC,

2 = , -5 =

∴x₁ = 1 and y₁ = -12

∴ coordinates of C are (1, -12)

For midpoint E of diagonal BD,

2 = , -5 =

∴ x₂ = 5 and y₂ = -10

∴ coordinates of D are (5, -10)

Let A(x₁ , y₁), B(x₂ , y₂) and C(x₃ , y₃) be the vertices of triangle.

Let D(3, 4), E(4, 6) and F(5, 7) be the midpoints of sides BC, CA and AB respectively.

By midpoint formula.

x = , y =

For midpoint D(3, 4) of side BC,

3 = , 4 =

∴ = 6 and = 8 …(1)

For midpoint E(4, 6) of side CA,

4 = , 6 =

∴ = 8 and = 12 …(2)

For midpoint F(5, 7) of side AB,

5 = , 7 =

∴ = 10 and = 14 …(3)

Adding 1,2 and 3, we get,

=6 + 8 + 10

And = 8 + 12 + 14

∴ 2(= 24 and 2() = 34

∴ = 12 and = 17

+ 6 = 12 and + 8 = 17 …from 1

∴ = 6 and = 9

Substituting above values in 3,

6 + and 9 + = 14

∴ = 4 and = 5

Similarly for equation 2,

6 + = 8 and 9 + = 12

∴ = 2 and = 3

Hence the vertices of triangle are A(6 , 9), B(4 ,5) and C(2 ,3)

Here, given points are P (3, 3) and Q (6, - 6) which is trisected at the points(say) A(x₁ , y₁) and B(x₂ , y₂)such that A is nearer to P.

By section formula,

x = , y =

For point A(x₁ , y₁) of PQ, where m = 2 and n = 1,

x₁ = , y₁ =

∴ x₁ = 4 , y₁ = 0

∴Coordinates of A is (4,0)

It is given that point A lies on the line 2x + y + k = 0.

So, substituting value of x and y as coordinates of A,

2 × 4 + 0 + k = 0

∴ k = - 8

Let given points be A(-2, -1), B(1, 0), C(x, 3), D(1, y) and let the intersection of diagonals be E(x_m , y_m )

It is given that □ABCD is a parallelogram.

By midpoint formula.

x = , y =

We know that midpoint of parallelogram coincide.

∴ Midpoint of AC = Midpoint of BD

∴ ( , ) = ( , )

∴ = and =

∴x = 4 and y = 2

Here given points are A (2, 0), B (9, 1), C (11, 6) and D (4, 4).

For a quadrilateral to be rhombus, all sides must be equal.

By distance formula,

XY =

For side AB,

AB =

=

= units.

For BC,

BC =

=

= units

CD =

=

= units.

AD =

=

= units.

Here all sides are unequal.

Hence □ABCD is not a rhombus.

Let three vertices be A(1, -2), B(3, 6) and C(5, 10) and fourth vertex be D(x, y)

It is given that quadrilateral joining these four vertices is parallelogram, ie □ABCD is parallelogram.

We know that diagonals of parallelogram bisect each other, ie midpoint of the diagonals coincide.

Let E(x_m , y_m) be the midpoint of diagonals AC and BD.

By midpoint formula,

x = , y =

For diagonal AC,

x_m = , y_m =

∴ x_m = , y_m =

∴ E(x_m , y_m) ≡ (3, 4)

For diagonal BD,

3 = , 4 =

∴ x = 6 – 3 , y = 8 – 6

∴ x = 3 and y = 2

Hence, our fourth vertex is D(3 , 2)

Given: the points A(a, -11), B(5, b), C(2, 15) and D(1, 1) are the vertices of a parallelogram ABCD.

To find: the values of a and b.

Solution:


Given points are A(a, -11), B(5, b), C(2, 15) and D(1, 1) and let the intersection of diagonals be E(x_m , y_m )

It is given that □ABCD is a parallelogram.

By midpoint formula.

x = , y =

We know that midpoint of parallelogram coincide.

∴ Midpoint of AC = Midpoint of BD

∴ ( , ) = ( , )

∴ = and =












⇒ a + 2 = 6 and 4 = b + 1

⇒ a = 6 - 2 and 4 - 1 = b

∴a = 4 and b = 3

Let A(x₁ , y₁), B(x₂ , y₂) and C(x₃ , y₃) be the vertices of triangle.

Let D(3, -2), E(-3, 1) and F(4, -3) be the midpoints of sides BC, CA and AB respectively.

By midpoint formula.

x = , y =

For midpoint D(3, -2) of side BC,

3 = , -2 =

∴ = 6 and = -4 …(1)

For midpoint E(-3, 1) of side CA,

-3 = , 1 =

∴ = -6 and = 2 …(2)

For midpoint F(4, -3) of side AB,

4 = , -3 =

∴ = 8 and =-6 …(3)

Adding 1,2 and 3, we get,

=6 -6 + 8

And = -4 + 2 -6

∴ 2(= 8 and 2() = -8

∴ = 4 and = -4

+ 6 = 4 and – 4 = -4 …from 1

∴ = -2 and = 0

Substituting above values in 3,

-2 + and 0 + = -6

∴ = 10 and = -6

Similarly for equation 2,

-2 + = -6 and 0 + = 2

∴ = -4 and = 2

Hence the vertices of triangle are A(-2 , 0), B(10 ,-6) and C(-4 ,2)

Here given vertices are A (0,-1), B (2, 1) and C (0, 3) and let midpoints of BC, CA and AB be D,E and F respectively.

By midpoint formula.

x = , y =

For midpoint D of side BC,

x = , y =

x = , y =

∴midpoint of side BC is D(1, 2)

For midpoint E of side AB,

x = , y =

x = , y =

∴midpoint of side AB is E(0, 1)

For midpoint F of side CA,

x = , y =

x = , y =

∴midpoint of side CA is F(1, 0)

By distance formula,

XY =

For median AD,

AD =

=

= units

For median BE,

BE =

=

= 2 units.

For median CF,

CF =

=

= units

Here given vertices are A (0,-1), B (2, 1) and C (0, 3) and let midpoints of BC, CA and AB be D,E and F respectively.

By midpoint formula.

x = , y =

For midpoint D of side BC,

x = , y =

x = , y =

∴midpoint of side BC is D(-1, 2)

For midpoint E of side AB,

x = , y =

x = , y =

∴midpoint of side AB is E(1, 0)

For midpoint F of side CA,

x = , y =

x = , y =

∴midpoint of side CA is F(3, 3)

By distance formula,

XY =

For median AD,

AD =

=

= units

For median BE,

BE =

=

= 5 units.

For median CF,

CF =

=

= units

Let given coordinates be A(-4, 0) and B (0, 6).

We need to divide AB into 4 equal parts, ie first we need to find midpoint of AB, which will be D and then find out midpoints of AD and DB respectively.

Let required points be C(x₁ , y₁ ), D(x_m , y_m ) and E(x , y₂ )

By midpoint formula.

x = , y =

For midpoint D of AB,

x_m = , y_m =

∴ x_m = -2 and y_m = 3

∴ D(x_m , y_m ) ≡ (-2 , 3)

Now, for midpoint C of AD,

X₁ = , y₂ =

x₁ = -3 and y₁ = 1.5

∴C(x₁ , y₁) ≡ (-3, 1.5)

For midpoint E of DB,

X₂ = , y₂ =

∴ x₂ = -1 and y₂ = 4.5

∴ D(x₂ , y₂ ) ≡ (-1 , 4.5)

Hence the co-ordinates of the points are (-3, 1.5), (-2 , 3) and (-1 , 4.5)

Let given points be A(5, 7) and B(3, 9) and the points of other segment line be C(8, 6) and D(0, 10)

By midpoint formula.

x = , y =

For midpoint of AB,

x = , y =

x = 4 and y = 8 …(1)

Now for midpoint of CD,

e = , d = …(say)

∴ e = 4 and d = 8 …(2)

Here from 1 and 2 we say that midpoints of AB and CD are same, ie they coincide.

Let D(x, y) be the midpoints of A(6, 8) and B(2, 4). Let our third given point be C(1, 2).

By midpoint formula.

x = , y =

For midpoint D of AB,

x = and y =

∴ x =4 and y = 6

∴D(x, y) ≡ (4, 6)

Now to find distance between C and D,

By distance formula,

XY =

For CD,

CD =

=

= 5 units

Given points are A(1, 4) and B(5, 2). Let P be (x, y) and given ratio is 3:4.

By section formula,

x = , y =

For point P on AB,

x = , y =

x = and y =

Hence, required coordinates is P( , )

Let given points be A (1, 0), B (5, 3), C (2, 7) and D (-2, 4) and let the intersection of diagonals be E(x_m , y_m )

By midpoint formula.

x = , y =

For midpoint of diagonal AC,

X₁ = , y₁ =

∴x₁ = , y_{1 =}

∴ midpoint of diagonal AC is (x₁, y₁ ) ≡ (, ) …(1)

For midpoint of diagonal BD,

X₂ = , y₂ =

∴x₂ = , y ₂₌

∴ midpoint of diagonal BD is (x₂, y₂ ) ≡ (, ) …(2)

Here, from 1 and 2 we say that midpoint of both the diagonals intersect at same point, ie (, )

But our intersection of diagonals is at E, which means that midpoint of diagonals intersect at single point, ie E(, )

We know that if midpoints of diagonals intersect at single point, then quadrilateral formed by joining the points is parallelogram.

Hence, our □ABCD is parallelogram.

Here, given points are A (-4, 3) and B ( 2, 8) and let the point dividing the line joining two points be P(m,6).

Let the ratio be m:n

By section formula,

x = , y =

For point P(m,6),

m = …(1)

And 6 = …(2)

Solving 2 for finding ratio between m and n,

6 =

6(m + n) = 8m +3n

6m + 6n = 8m +3n

∴ 2m = 3n

∴ =

∴ m : n = 3 : 2

Now solving for equation 1, where m = 3 and n =2

m =

∴ m =

∴ m =

Hence, our point is (, 6)

Here, given points are A(-3, 1) and B(-8, 9) and let the point dividing the line joining two points be C(-6,a).

Let the ratio be m:n

By section formula,

x = , y =

For point C(-6,a),

-6 = …(1)

And a = …(2)

Solving 1 for finding ratio between m and n,

-6 =

-6 (m + n) = -8m -3n

6m + 6n = 8m + 3n

∴ 2m = 3n

∴ =

∴ m : n = 3 : 2

Now solving for equation 2, where m = 3 and n = 2

a =

a =

∴ a =

∴ a =

∴ value of a is

Let given points be A(3, -4) and B(1, 2) , which is trisected at points P(p, -2) and Q(5/3, q).

By section formula,

x = , y =
As point P divides the line in 1:2 and Q divides the line in 2:1.

For point P(p, -2)of AB, where m = 1 and n = 2,

p= ,
-2 =

Solving for p,

P =

For point Q(5/3, q) of AB, where m = 2 and n = 1,

= ,
q =

Solving for q,

q =

∴q = 0
Hence, the value of p and q are and 0 respectively.

Here, given points are P (2, 1) and Q (5, - 8) which is trisected at the points(say) A(x₁ , y₁) and B(x₂ , y₂)such that A is nearer to P.

By section formula,

x = , y =

For point A(x₁ , y₁) of PQ, where m = 1 and n = 2,

x₁ = , y₁ =

∴ x₁ = 3 , y₁ = -2

∴Coordinates of A is (3,-2)

It is given that point A lies on the line 2x - y + k = 0.

So, substituting value of x and y as coordinates of A,

2 × 3 – (-2) + k = 0

∴ k = -8

Given points are A(-2, -2) and B(2, -4). Let P be (x, y)

Here given that AP = AB.

But AB = AP + BP

∴7AP = 3AB

7AP = 3(AP + BP)

∴4AP = 3BP

∴ =

By section formula,

x = , y =

For point P on AB, where m = 3 and n = 4

x = , y =

x = and y =

Hence, required coordinates is P( , )

Let given coordinates be A(-2, 2) and B (2, 8).

We need to divide AB into 4 equal parts, ie first we need to find midpoint of AB, which will be D and then find out midpoints of AD and DB respectively.

Let required points be C(x₁ , y₁ ), D(x_m , y_m ) and E(x , y₂ )

By midpoint formula.

x = , y =

For midpoint D of AB,

x_m = , y_m =

∴ x_m = 0 and y_m = 5

∴ D(x_m , y_m ) ≡ (0 , 5)

Now, for midpoint C of AD,

X₁ = , y₂ =

x₁ = -1 and y₁ =

∴C(x₁ , y₁) ≡ (-1, )

For midpoint E of DB,

X₂ = , y₂ =

∴ x₂ = 1 and y₂ =

∴ E(x₂ , y₂ ) ≡ (1 , )

Hence the co-ordinates of the points are (-1, ) , (0 , 5) and (1 , )

(i) The median from A meets BC in D. Find the coordinates of the point D.

Here given vertices are A (4, 2), B (6, 5) and C (1, 4).

By midpoint formula.

x = , y =

For midpoint D of side BC,

x = , y =

x = , y =

Hence, the coordinates of D are ( , )

(ii) Find the coordinates of point P on AD such that AP : PD = 2 :1.

By section formula,

x = , y =

For point P on AD, where m = 2 and n = 1

x = , y =

∴ x = and y =

(iii) Find the coordinates of the points Q and R on medians BE and CF respectively such that BQ : QE = 2 : 1 and CR : RF = 2 : 1.

By midpoint formula.

x = , y =

For midpoint E of side AC,

x = , y =

x = , y =

Hence, the coordinates of E are ( , 3)

For midpoint F of side AB,

x = , y =

x = , y =

Hence, the coordinates of F are ( , )

By section formula,

x = , y =

For point Q on BE, where m = 2 and n = 1

x = , y =

∴ x = and y =

For point R on CF, where m = 2 and n = 1

x = , y =

∴ x = and y =

(iv) What do you observe?

We observe that the point P,Q and R coincides with the centroid.

This also shows that centroid divides the median in the ratio 2:1

Here given that A (-1, -1), B (-1, 4), C (5, 4) and D (5,-1).Also P, Q, R and S are the mid-points of sides AB, BC, CD and DA respectively.

By midpoint formula.

x = , y =

For midpoint P of side AB,

x = , y =

x = -1 , y =

Hence, the coordinates of P are (-1 , )

For midpoint Q of side BC,

x = , y =

x = 2 , y = 4

Hence, the coordinates of Q are (2 ,4)

For midpoint R of side CD,

x = , y =

x = 5 , y =

Hence, the coordinates of R are (5 , )

For midpoint S of side AD,

x = , y =

x = 2 , y = -1

Hence, the coordinates of S are (2 ,-1)

Now we find length of the length of the □PQRS,

By distance formula,

XY =

For PQ,

PQ =

=

= units

For QR,

QR =

=

= units

For RS,

RS =

=

= units

For PS,

PS =

=

= units

Here we can observe that all lengths of □PQRS are equal.

Now for diagonal PR,

PR =

=

= units

Now for diagonal QS,

QS =

=

= 5 units

Here in □PQRS, diagonals are unequal.

We know that a quadrilateral whose all sides are equal and diagonals are unequal, it is a rhombus.

Hence, our □PQRS is rhombus .

solution: Given points are A(-3, 2), B (-5, -5), C (2, -3) and D(4, 4)

Use distance formula

For AB ,

AB =

=

= units

For BC ,

BC =

=

= units

For CD,

CD =

=

= units

For AD ,

AD=

=

= units

Here we can observe that all lengths of □PQRS are equal.

Now for diagonal AC,

AC =

=

= units

Now for diagonal BD,

BD =

=

= units


AB = BC = CD = AD

Here in ABCD, diagonals are unequal.

We know that a quadrilateral whose all sides are equal and diagonals are unequal, it is a rhombus.

Hence, ABCD is rhombus .

Let our points be A(5, -6) and B(-1, -4).

Let point C(0, y) divide the line formed by joining by the points A and B in ratio of m:n.

By section formula,

x = , y =

For point C(0, y)

0 = , y =

Solving for x coordinate,

0 =

∴ m = 5n

∴ =

∴ m : n = 5 : 1

Now solving for y coordinate, with m = 5 and n = 1,

y =

y =

∴ y =

There is no need to solve for x, as our point lies on y-axis

Hence, the coordinates of required point is C( , )

Our given vertices are A(1, -2), B(3, 6) and C(5, 10) and fourth vertex be D(k, p)

It is given that quadrilateral joining these four vertices is parallelogram, ie □ABCD is parallelogram.

We know that diagonals of parallelogram bisect each other, ie midpoint of the diagonals coincide.

Let E(x_m , y_m) be the midpoint of diagonals AC and BD.

By midpoint formula,

x = , y =

For diagonal AC,

x_m = , y_m =

∴ x_m = , y_m =

∴ E(x_m , y_m) ≡ (, )

For diagonal BD,

= , =

∴ k = 15 – 8 , y = 5 – 2

∴ k = 7 and p = 3

Hence, our fourth vertex is D(7 , 3)

Given points are A (-6, 10) and B(3, -8)

Let the point C(-4, 6) divide AB in ratio m:n.

By section formula,

x = , y =

For point C(-4, 6) on the line joined by the points A and B.

-4 = …(1)

And, 6 = …(2)

Solving 1,

-4(m + n) = 3m - 6n

∴4m + 4n = -3m + 6n

∴ 7m = 2n

∴ =

Hence, ratio is 2:7.

Here given that AB is a diameter of the circle whose centre is (say) C(2, -3) and B is (1, 4)

Let A be (x, y)

We know that as C is center, AC = CB or C is midpoint of AB.

By midpoint formula,

x = , y =

For Center C,

2 = and -3 =

∴ x = 4 – 1 and y = -6 – 4

∴ x = 3 and y = -10

Hence, coordinates of A are (3, -10)

Here given points are A (3, -5) and B (-4, 8) .

Let point P be (x, y) which divides AB in ratio of k:1, also point P lies on line x + y = 0

By section formula,

x = , y =

For point P on the line joined by the points A and B.

x = , y =

Putting in given equation,

, = (x, y)

∴ 4k = 5

∴ k = 5/2

Here, given points are A ( -3,10) and B(6, -8) and the point dividing the line joining two points is P(-1,y).

Let the ratio be m:n

By section formula,

x = , y =

For point P(-1,a),

-1 = …(1)

And y = …(2)

Solving 1 for finding ratio between m and n,

-1 =

- (m + n) = 6m -3n

m + n = -6m + 3n

∴ 7m = 2n

∴ =

∴ m : n = 2 : 7

Now solving for equation 2, where m = 2 and n = 7

y =

y =

∴ y =

∴ y =

∴ value of y is

Here given points are A (1, 2) and B (6, 7) which is divided into 5 equal parts by points P, Q, R and S

∴ AP = PQ= QR = RS = SB

The point P divides the line segment AB in the ratio 1:4.

By section formula,

x = , y =

For point P,

x = , y =

x = 2 and y = 3

∴ Coordinate of P is (2 ,3 )

The point Q divides the line segment AB in the ratio of 2:3.

For point Q,

x = , y =

x = 3 and y = 4

∴ Coordinate of Q is (3 ,4 )

The point R divides the line segment AB in the ratio of 3:2.

For point R,

x = , y =

x = 4 and y = 5

∴ Coordinate of R is (4 , 5 )

Here given points are A (- 10, 4) and B (- 2, 0)and the points of other segment line are C (- 9,-4) and D (- 4, y)

Let the point of intersection between AB and CD be P

By midpoint formula.

x = , y =

For midpoint of AB,

e = , d = …(say)

e = -6 and d = 2 …(1)

By section formula,

x = , y =

For point P on CD, where ratio is m:n,

-6 = and 2 =

Solving for m and n,

-6 =

∴-6(m + n) = -4m -9n

6m + 6n = 4m + 9n

2m = 3n

∴ =

∴ Ratio is 3:2

Now solving for y, where m = 3 and n = 2,

2 =

∴ 3y -8 = 10

∴ 3y = 18

∴y = 6

Here, given points are A (12,5) and B (4, -3) and let the point dividing the line joining two points be P(x,2)

Let the ratio be m:n

By section formula,

x = , y =

For point P(x,2),

x = …(1)

And 2 = …(2)

Solving 2 for finding ratio between m and n,

2 =

2(m + n) = -3m +5n

2m + 2n = -3m +5n

∴ 5m = 3n

∴ =

∴ m : n = 3 : 5

Now solving for equation 1, where m = 3 and n =5

x =

∴ x =

∴ x = 9

Hence, our point is (9, 2)

Our points are A (3,-3) and B (-2, 7)

Let point C(x, 0) divide the line formed by joining by the points A and B in ratio of m:n.

By section formula,

x = , y =

For point C(x, 0)

x = , 0 =

Solving for y coordinate,

0 =

∴ 7m -3n = 0

∴ 7m = 3n

∴ =

∴ m : n = 3 : 7

Now solving for x coordinate, with m = 3 and n = 7,

x =

∴ x=

∴ x = =

Hence, the coordinates of required point is C(, 0)

Given points are A(1/2 , 3/2) and B(2, -5)

Let the point P(3/4, 5/12) divide AB in ratio m:n.

By section formula,

x = , y =

For point P on the line joined by the points A and B.

= …(1)

And, = …(2)

Solving 1,

3(m + n) = 8m + 2n

∴3m + 3n = 8m + 2n

∴ 5m = n

∴ =

Hence, ratio is 1:5.

Here given points are A (2, p) and B ( 7, 10) which is divided into 5 equal parts by points P, Q(x, 7), R and S(6, y)

∴ AP = PQ= QR = RS = SB

By section formula,

x = , y =

The point Q divides the line segment AB in the ratio of 2:3.

For point Q,

x = , 7 =

Solving above equations, we get,

x = 4 and p = 5

For point P, divides the line segment AB in the ratio 4:1.

6 = , y =

Solving for y and substituting value of p,

y =

∴ y = 9

Hence, values are x = 4, y = 9 and p = 5

(i) (1, 4), (-1, -1), (3, -2)

We know that centroid of a triangle for the vertices (x₁ , y₁), (x₂ , y₂) and (x₃ , y₃) is

G(x, y) = ( , )

∴For coordinates (1, 4), (-1, -1), (3, -2),

Centroid of triangle = , )

= ( 1, )

Hence, centroid of triangle is ( 1, )

(ii) (- 2, 3), (2, -1), (4, 0)

We know that centroid of a triangle for (x₁ , y₁), (x₂ , y₂) and (x₃ , y₃) is

G(x, y) = ( , )

∴For coordinates (- 2, 3), (2, -1), (4, 0)

Centroid of triangle = , )

= ( , )

Hence, centroid of triangle is ( , )

Let the vertex of the triangle be A(1, 2), B(3, 5) and C(x, y)

Let the centroid be D(0, 0), as it is given that centroid is given at origin.

We know that centroid of a triangle for (x₁ , y₁), (x₂ , y₂) and (x₃ , y₃) is

C(x, y) = ( , )

For given coordinates A(1, 2), B(3, 5) and C(x, y), centroid is,

(0, 0) = , )

Solving for x and y,

1 + 3 +x =0 and = 0

∴ x = -4 and y = -7

Hence, the coordinate of third vertex is C(-4, -7)

Let ∆ABC be any triangle such that O is the origin.

∴Let coordinates be A(0, 0), B(x₁ , y₁), C(x₂ , y₂).

Let D and E are the mid-points of the sides AB and AC respectively.

We have to prove that line joining the mid-point of any two sides of a triangle is equal to half of the third side which means,

DE = BC

By midpoint formula,

x = , y =

For midpoint D on AB,

x =, y =

∴ x = and y =

∴ Coordinate of D is (, )

For midpoint E on AC,

x =, y =

∴ x = and y =

∴ Coordinate of E is ( , )

By distance formula,

XY =

For BC,

BC =

For DE,

DE =

= ( )

= BC

∴ DE = BC

Hence, we proved that line joining the mid-point of any two sides of a triangle is equal to half of the third side.

Let us consider a Cartesian plane having a parallelogram OABC in which O is the origin.

We have to prove that middle point of the opposite sides of a quadrilateral and the join of the mid-points of its diagonals meet in a point and bisect each other.

Let coordinates be A(0, 0).

So other coordinates will be B(x_{1 +} x₂, y₁), C(x₂, 0) ... refer figure.

Let P, Q, R and S be the mid-points of the sides AB, BC, CD, DA respectively.

By midpoint formula,

x = , y =

For midpoint P on AB,

x =, y =

∴ x = , y =

∴ Coordinate of P is ( , )

For midpoint Q on BC,

x =, y =

∴ x = , y =

∴ Coordinate of Q is ( , )

For R, we can observe that, R lies on x axis.

∴ Coordinate of R is ( , )

For midpoint S on OA,

x =, y =

∴ x = , y =

∴ Coordinate of S is ( , )

For midpoint of PR,

x = , y =

∴ x = , y =

∴ Midpoint of PR is ( , )

Similarly midpoint of QS is ( , )

Also, similarly midpoint of AC and OA is ( , )

Hence, midpoints of PR, QS, AC and OA coincide

∴We say that middle point of the opposite sides of a quadrilateral and the join of the mid-points of its diagonals meet in a point and bisect each other.

we will solve it by taking the coordinates A(x₁,y₁), B(x₂,y₂) and C(x₃,y₃)

Let the co ordinates of the centroid be G(u, v).

G(u, v) = ( , )

let the coordinates of P(h, k).

now we will find L.H.S and R.H.S. separately.

PA²+PB²+PC²

= (h - x₁)² +(k - y₁)² +(h - x₂)²+ (k - y₂)² +(h - x₃)² +(k - y₃)² …by distance formula.

= 3(h²+k²)+(x₁²+x₂²+x₃²)+(y₁²+y₂²+y₃²)-2h(x₁+x₂+x₃)-2k(y₁+y₂+y₃)

= 3(h²+k²)+(x₁²+x₂²+x₃²)+(y₁²+y₂²+y₃²)-2h(3u)-2k(3v)

GA²+GB²+GC²+3GP²

= (u-x₁)²+(v-y₁)²+(u-x₂)²+(v-y₂)²+(u-x₃)²+(v-y₃)²+3[(u-h)²+(v-k)²] ……by distance formula.

= 3(u²+v²)+(x₁²+y₁²+x₂²+y₂²+x₃²+y₃²) - 2u(x₁+x₂+x₃) - 2v(y₁+y₂+y₃) + 3[u²+h²-2uh+v²+k²-2vk]

= 6(u²+v²)+(x₁²+y₁²+x₂²+y₂²+x₃²+y₃²)-2u(3u)-2v(3v) +3(h²+k²) -6uh-6vk

= (x₁²+x₂²+x₃²)+(y₁²+y₂²+y₃²)+3(h²+k²)-6uh-6vk

Hence LHS = RHS

(The above relation is known as Leibniz Relation)

Hence Proved.

We know that centroid of a triangle for (x₁ , y₁), (x₂ , y₂) and (x₃ , y₃) is

G(x, y) = ( , )

We assume centroid of ∆ABC at origin.

For x=0 and y=0

= 0 and = 0

∴ = 0 and =0

Squaring on both sides, we get

x₁² + x₂² + x₃² + 2x₁x₂ + 2x₂x₃ + 2x₃x₁ = 0 and y₁² + y₂² + y₃² + 2y₁y₂ + 2y₂y₃ + 2y₃y₁ = 0 … (1)

AB² + BC² + CA²

= [(x₂ – x₁)² + (y₂ – y₁)²] + [(x₃ – x₂)² + (y₃ – y₂)²] + [(x₁ – x₃)² + (y₁ – y₃)²]

= (x₁² + x₂² – 2x₁x₂ + y₁² + y₂² – 2y₁y₂)+(x₂² + x₃² – 2x₂x₃ + y₂² + y₃² – 2y₂y₃)+(x₁² + x₃² – 2x₁x₃ + y₁² + y₃² - 2y₁y₃)

= (2x₁² + 2x₂² + 2x₃² – 2x₁x₂ – 2x₂x₃ – 2x₁x₃) + (2y₁² + 2y₂² + 2y₃² – 2y₁y₂ – 2y₂y₃ – 2y₁y₃)

= (3x₁² + 3x₂² + 3x₃²) + (3y₁² + 3y₂² + 3y₃²) …from 1

= 3(x₁² + x₂² + x₃²) + 3(y₁² + y₂² + y₃²) … (2)

3(GA² + GB² + GC²)

= 3 [(x₁ – 0)² + (y₁ – 0)² + (x₂ – 0)² + (y₂ – 0)² + (x₃ – 0)² + (y₃ – 0)²]

= 3 (x₁² + y₁² + x₂² + y₂² + x₃² + y₃²)

= 3 (x₁² + x₂² + x₃²) + 3(y₁² + y₂² + y₃²) … (3)

From (2) and (3), we get

AB² + BC² + CA² = 3(GA² + GB² + GC²)

We know that centroid of ∆DEF will be the same that of ∆ABC as ∆DEF is formed by midpoints of ∆ABC.

∴ We know that centroid of a triangle for (x₁ , y₁), (x₂ , y₂) and (x₃ , y₃) is

G(x, y) = ( , )

∴ G(x, y) = ( , )

∴ G(x, y) = ( 2, )

Hence the centroid is ( 2, )

Given that ∆BOA is right angled triangle

By midpoint formula,

x = , y =

For midpoint C on AB,

x =, y =

∴ x = a and y = b

∴ Coordinates of C are (a, b)

It is given that C is the midpoint of AB.

By distance formula,

XY =

For OC,

OC =

= …(1)

For AC,

AC =

=

As C is midpoint, AC = CB. …(2)

Hence from 1 and 2, we say that is point C is equidistant from the vertices 0, A and B.

Let the vertex of the triangle be A(1, 2), B(3, 5) and C(x, y)

Let the centroid be G(0, 0), as it is given that centroid is given at origin.

We know that centroid of a triangle for (x₁ , y₁), (x₂ , y₂) and (x₃ , y₃) is

G(x, y) = ( , )

For given coordinates A(1, 2), B(3, 5) and C(x, y), centroid is,

(0, 0) = , )

Solving for x and y,

-3 +x =0 and = 0

∴ x = 3 and y = 1

Hence, the coordinate of third vertex is C(3, 1).

Let the vertex of the triangle be A(3, 2), B(-2, 1) and C(x, y)

Let the centroid be G(, ), as it is given that centroid is given at origin.

We know that centroid of a triangle for (x₁ , y₁), (x₂ , y₂) and (x₃ , y₃) is

G(x, y) = ( , )

For given coordinates A(3, 2), B(-2, 1) and C(x, y)

(, ) = , )

Solving for x and y,

-3 + 2 +x =5 and 2 + = -1

∴ x = 6 and y = -4

Hence, the coordinate of third vertex is C(6, -4).

(i) (6, 3), (-3, 5) and (4, - 2)

Let A≡(x₁, y₁ ) ≡ (6, 3), B ≡ (x₂, y₂ ) ≡ (-3, 5) and C ≡ (x₃, y₃ ) ≡ (4, -2)

Area of ∆ABC = |x₁ (y₂ - y₃) + x₂ (y₃ - y₁) + x₃ (y₁ - y₂)| sq. units

∴ Area of ∆ABC = | { 6 (5 – (-2)) – 3 ( -2 – 3) + 4 (3 – 5)} |

= | { 6 × 7 + 15 – 8 } |

= | 57 – 8|

= sq. units

(ii)

Area of the triangle having vertices (x₁,y₁), (x₂,y₂) and (x₃,y₃)

= |x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)|

Here, (x₁, y₁)=(at₁²,2at₁), (x₂,y₂)=(at₂²,2at₂), (x₃,y₃)=(at₃²,2at₃)

∴, area=|at₁²(2at₂-2at₃)+at₂²(2at₃-2at₁)+at₃²(2at₁-2at₂)|

= |2a²t₁²t₂-2a²t₁²t₃+2a²t₂²t₃-2a²t₂²t₁+2a²t₃²t₁-2a²t₃²t₂|

= × 2a²|t₁²t₂-t₁²t₃+t₂²t₃-t₂²t₁+t₃²t₁-t₃²t₂|

=a²|t₁²t₂-t₁²t₃+t₂²t₃-t₂²t₁+t₃²t₁-t₃²t₂|

=a²|t₁²(t₂-t₃)+t₂t₃(t₂-t₃)-t₁(t₂²-t₃²)|

=a²|t₁²(t₂-t₃)+t₂t₃(t₂-t₃)-t₁(t₂+t₃)(t₂-t₃)|

=a²|(t₂-t₃)(t₁²+t₂t₃-t₁t₂-t₁t₃)|

=a²|(t₂-t₃){t₁(t₁-t₂)-t₃(t₁-t₂)}|

=a²|(t₂-t₃)(t₁-t₂)(t₁-t₃)|

∴ Area is a²|(t₂-t₃)(t₁-t₂)(t₁-t₃)| sq. units