Coordinate Geometry

(x_1,y₁) = (2,3) and (x_2,y₂) = (4,1)

d =

=

=

=

d = units

(x_1,y₁) = (-5,7) and (x_2,y₂) = (-1,3)

d =

=

=

=

d =

(x_1,y₁) = (-2,-3) and (x_2,y₂) = (3,2)

d =

=

=

d =

(x_1,y₁) = (a,b) and (x_2,y₂) = (-a,-b)

d =

=

=

d =

let (x_1,y₁) = (0,0) and (x_2,y₂) = (36,15)

D =

=

=

=

d =

let A = (1,5) B = (2,3) and c = (-2,-1)

⇒ AB =

=

AB =

⇒ BC =

BC =

⇒ AC =

=

AC =

⇒ AB + BC is not equal to AC.

∴ Points are not collinear

let A = (5,-2) B = (6,4) and c = (7,2)

⇒ AB =

=

AB =

⇒ BC =

=

BC =

⇒ AC =

=

AC =

As all the sides are unequal the triangle is not isosceles.

let A = (3,4) B = (6,7) and C = (9,4) D = (6,1)

⇒ AB =

=

AB =

⇒ BC =

=

BC =

⇒ CD =

=

CD =

⇒ AD =

=

AD =

As all the sides are equal the ABCD is a square. Jarina is correct.

let A = (a,0) B = (-a,0) and

⇒ AB =

=

AB = 2a

⇒ BC =

=

=

BC = 2a

⇒ AC =

=

=

AC = 2a

As all the sides are equal the triangle is Equilateral.

let A = (-7,-3) B = (5,10) and C = (15,8) D = (3,-5)

Let these points be a parallelogram.

So midpoints of AC and DB should be same.

⇒ To find midpoint of AC and DB

⇒ For AC =

AC =

AC =

⇒ For DB

DB =

DB =

DB =

As midpoints of AC and DB are same the points form a parallelogram.

Let us divide the parallelogram into 2 triangle ΔABD and ΔBCD

Area of both triangles

⇒ Area ΔABD =

=

=

=

=

= 77

⇒ Area ΔBCD =

=

=

=

= 77

Total area of parallelogram = Sum of Area of triangles

= ΔBCD + ΔABD

= 154 units

Let the points be

A(-4, -7), B(-1, 2), C(8, 5) and D(5, -4)

Length of diagonals

AC =

AC =

AC =

BD =

BD =

BD =

⇒ Area of Rhombus = 1/2 × Product of diagonals

= =

= 72 units

⇒ Area of triangles

Area ΔABD =

=

=

=

= 36

⇒ Are

a ΔBCD =

=

=

=

= 36

Sum of area of triangles = 36 + 36 = 72 units

Thus proved

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

AB =

AB =

AB =

AB =

BC =

BC =

BC =

BC =

CD =

CD =

CD =

AD =

AD =

AD =

As all sides are equal the quadrilateral is a square

Let A(-3, 5), B(3, 1), C(1, -3) and D(-1,-4)

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

If ABCD is parallelogram

Midpoint of diagonals AC and BD

⇒ For AC =

X(4,4)

⇒ For BD =

X(4,4)

As the midpoints are same the diagonals bisect each other

Thus, the points form a parallelogram

Let P(x,0) be the point

A(2,-5) and B(-2,9)

⇒ PA =

⇒ PB =

PA = PB

PA² = PB²

(x-2)² + (-5)² = (x + 2)² + 9²

x²-4x + 4 + 25 = x² + 4x + 4 + 81

8x = -56

X = 7

Point is (-7,0)

7 or -5

Let A(x,7) and B(1,15)

be the point

⇒ AB =

AB = 10

AB² = 10²

⇒ (1-x)² + 8² = 10²

⇒ x²-2x + 1 + 64 = 100

⇒ x²-2x-35 = 0

⇒ (x-7)(x + 5) = 0

⇒ X = 7 or x = -5

3 or -9

Let P(2,-3) and Q(10,y)

be the point

⇒ PQ =

⇒ PQ = 10

PQ² = 10²

⇒ (y + 3)² + 8² = 10²

y² + 6y + 9 + 64 = 100

y² + 6y-27 = 0

⇒ (y + 9)(y-3) = 0

⇒ y = 9 or y = -3

Let P be center such that P(3,2) and Q be point on circumference Q(-5,6)

PQ =

PQ =

PQ =

PQ =

PQ =

Let A(1, 5), B(5, 8) and C(13, 14)

⇒ AB =

AB =

AB = 5

⇒ BC =

BC =

BC = 10

⇒ AC =

AC =

AC = 15

Largest side is AC = 15 which is equal to Sum of other two sides AB and BC i.e. (10 + 5 = 15)

∴ triangle cannot be drawn infact the points are collinear

Let P(x,y) be the point

A(-2,8) and B(-3,-5)

⇒ PA =

⇒ PB =

PA = PB

PA² = PB²

(x + 2)² + (8-y)² = (x + 3)² + (y + 5)²

x² + 4x + 4 + 64 -16y + y² = x² + 6x + 9 + y² + 10y + 25

-2x-26y = -34

⇒ X + 13y = 17

Let P(x,y) be the point

P(x,y) = (1,3)

points A(4, -1) and B(-2, -3) are shown in the graph above
Now points of trisection means the points which divides the line in three parts. From the figure it is clear that, there will be 2 points which will do that. Let us call them P and Q.
Now clearly point P divides the BA in the ratio 1:2 and point Q divides the line in the ratio 2:1
⇒ Let P and Q be points of trisection

∴ P divides BA internally in ratio 1:2

Such that AP = PQ = QB

⇒ Q divides BA internally in ratio 2:1

Hence, Points of trisection are and

Let P and Q be line

∴ A divides PQ internally in ratio a:b

⇒ Given that A(-1,6)

6a-3b = -a-b , 10b-8a = 6a + 6b

7a = 2b, 14a = 4b

∴ a:b = 2:7

Let A(1, 2), B(4, y), C(x, 6) and D(3, 5)

If ABCD is a parallelogram

AC and BD bisect each other

⇒ Midpoint of AC

⇒ For AB =

=

⇒ Midpoint of BD

For BD =

=

∴

1 + x = 7 , 8 = 5 + y

x = 6 y = 3

Let O be the center O(2,-3)

⇒ A(x,y) B(1,4)

∴ O divides AB internally in ratio 1:1

x + 1 = 4

x = 3

y + 4 = -6

y = -10

AP = 3/7AB

∴ P divides AB in ratio 3:4

⇒ Let x,y and z divide the line into 4 equal parts such that

AX = XY = YZ = ZB

⇒ X divides AB in ration 1:3

⇒ Y divides AB in ratio 1:1

Y(-2,3)

⇒ Z divides AB in ratio 3:1

⇒ Let x,y and z divide the line into 4 equal parts such that

AX = XY = YZ = ZB

⇒ X divides AB in ration 1:3

⇒ Y divides AB in ratio 1:1

Y(0,5)

⇒ Z divides AB in ratio 3:1

The coordinates of centroid are

Area ΔABC =

=

=

=

Area ΔABC =

=

=

=

= 32

Area ΔABC =

=

=

=

= 3

For points to be collinear area of ΔABC = 0

Area ΔABC =

=

=

=

=

∴ 8 - 2k = 0

K = 4

For points to be collinear area of ΔABC = 0

Area ΔABC =

=

=

=

18-6k = 0

K = 3

For points to be collinear area of ΔABC = 0

Area ΔABC =

=

=

=

=

⇒ K² -11k-12 = 0

⇒ K(k-12) + (k-12) = 0

⇒ (K-12)(k + 1) = 0

⇒ K = 12 or k = -1

1 sq. unit; 1 : 4

⇒ Let A(0,-1),B(2,1),C(0,3)

So midpoints of AB and BC and CD

⇒ Midpoint formula

⇒ For AB =

AB =

AB =

⇒ For BC

BC =

BC =

BC =

⇒ For AC

AC =

AC =

AC =

(1,0)(1,2)(0,1)

Area ΔXYZ =

=

=

=

= 1 sq cm

⇒ Let A(0,-1),B(2,1),C(0,3)

⇒ Area ΔABC =

=

=

=

= 4 sq cm

∴ Ratio of Area of triangles is 1:4

28 sq. units

⇒ Let A = (-4, -2), B = (-3, -5), C = (3, -2) and D = (2, 3)

We draw Line BD and divide the quadrilateral into 2 triangles

ΔABD and ΔBDC

Area of both triangles

⇒ Area ΔBDC =

=

=

=

=

=

= 16.5 units

A = (-4, -2), B = (-3, -5), D = (2, 3)

Area ΔABD =

=

=

=

=

= 11.5 units

⇒ Total area of quadrilateral = Sum of Area of triangles

= ΔBCD + ΔABD

= 16.5 + 11.5 units

= 28 sq units.

Not possible

Let A = (8,-5) B = (-2,-7) and C = (5,1)

AB =

AB =

AB =

BC =

BC =

BC =

AC =

AC =

AC =

Slope