Geometry And Algebra

Let the coordinate of fourth vertex be (x₁, y₁) and the point of intersection of diagonals be (x₂, y₂)

Diagonals of a parallelogram bisects each other

According to the section formula for mid points

⇒ x₂ = 3.5

y₂

⇒ y₂ = 4.5

The point of intersection is (3.5, 4.5)

Again using the section formula for mid points

⇒ 1 + x₁ = 7

⇒ x₁ = 6

⇒ 1 + y₁ = 9

⇒ y₁ = 8

The fourth vertex of parallelogram is (6, 8)

Let the three coordinates of the larger triangle be (x₁, y₁), (x₂, y₂) and (x₃, y₃)

The three mid points on the three sides are (3, 3) , (4, 2) , (5, 4)

According to the section formula for mid points

⇒ x₁ + x₂ = 6 …Equation(i)

⇒ x₂ + x₃ = 8 …Equation (ii)

⇒ x₃ + x₁ = 10 …Equation(iii)

Solving Equation (i) ,(ii) and (iii)

x₂ = 6 - x₁

Putting this value in equation (ii) we get

x₃ - x₁ = 2 …Equation (iv)

Solving Equation (iii) & (iv)

2x₃ = 12

⇒ x₃ = 6

Putting in Equation (iv)

6 - x₁ = 2

⇒ x₁ = 4

Putting this value in equation (i)

4 + x₂ = 6

⇒ x₂ = 2

⇒ y₁ + y₂ = 6 …Equation (v)

⇒ y₂ + y₃ = 4 …Equation (vi)

⇒ y₃ + y₁ = 8 …Equation (vii)

Solving Equation (v) ,(vi) and (vii)

y₂ = 6 - y₁

Putting this value in equation (vi) we get

y₃ - y₁ = - 2 …Equation (viii)

Solving Equation (vii) & (viii)

2y₃ = 6

⇒ y₃ = 3

Putting in Equation (viii)

3 - y₁ = - 2

⇒ y₁ = 5

Putting this value in equation (v)

5 + y₂ = 6

⇒ y₂ = 1

The Vertices of the larger triangle are (4,5) , (2,1) , (6,3)

Let the coordinate of fourth vertex be (a₁, b₁) and the point of intersection of diagonals be (a₂, b₂)

Diagonals of a parallelogram bisects each other

According to the section formula for mid points

a₂

⇒ a₂ = 0.5(x₁ + x₂)

b₂

⇒ b₂ = 0.5(y₁ + y₂)

The point of intersection is (0.5(x₁ + x₂), 0.5(y₁ + y₂))

Again using the section formula for mid points

⇒ a₁ = x₁ + x₂

⇒ b₁ = y₁ + y₂

The fourth vertex of parallelogram is (x₁ + x₂, y₁ + y₂)

In the parallelogram ABCD, AB = CD & AD = BC

Let DF and CE be two perpendiculars drawn on AB

In ΔAEC using Pythagoras theorem we can say

AC² = AE² + CE²

Since AE = AB + BE so we can say

⇒ AC² = (AB + BE)² + CE²

AC² = AB² + BE² + 2 × AB × BE + CE² …Equation (i)

In ΔDBF using Pythagoras theorem we can say

DB² = DF² + BF²

Since BF = AB - AF so we can say

⇒ DB² = (AB - AF)² + DF²

DB² = AB² + AF² - 2 × AB × AF + DF² …Equation (ii)

In ΔDAF & ΔCBE

DA = CB (Opposite sides of a parallelogram)

DF = CE (DCEF is a rectangle)

∠DFA = ∠ CEB (Perpendiculars)

So ΔDAF & ΔCBE are congruent by S.A.S. axiom of congruency

AF = BE (Corresponding Parts of Congruent Triangle)

Adding Equation (i) and (ii)

AC² + DB² = AB² + AF² - 2 × AB × AF + DF² + AB² + BE² + 2 × AB × BE + CE²

⇒ AC² + DB² = AB² + AF² + DF² + AB² + BE² + CE²

(Since AF = BE)

Since AB = CD (Opposite side of parallelogram)

⇒ AC² + DB² = AB² + AF² + DF² + CD² + BE² + CE²

Using Pythagoras theorem

⇒ AC² + DB² = AB² + AD² + CD² + BC²

Hence Proved

(i) The points are A(3,2) , B(8,7)

AP : PB = 2 : 3(Given)

P(x,y) = (5,4)

(ii) The points are A(3,2) , B(8,7)

AQ : QB = 3 : 2(Given)

Q(x,y) = (6,5)

Let the mid points be A, B, C, D

Using the section formula for mid points

A(x, y)

A(x, y)

A(x, y) = (3.5,2)

B(x, y)

B(x, y)

B(x, y) = (6.5,5)

C(x, y)

C(x, y)

C(x, y) = (6,8)

D(x, y)

D(x, y)

D(x, y) = (3,5)

(ii) Length AB = √((6.5 - 3.5)² + (5 - 2)²)

Length AB = 3√2 units

Length BC = √((6.5 - 6)² + (5 - 8)²)

Length BC = √9.25 units

Length CD = √((6 - 3)² + (8 - 5)²)

Length CD = 3√2 units

Length DA = √((3.5 - 3)² + (2 - 5)²)

Length DA = √9.25 units

Since the lengths of opposite sides are equal hence it forms a parallelogram.

Hence the quadrilateral forms a parallelogram by joining the mid points .

Let the three vertices of larger quadrilateral be A, B ,C

⇒ A(x) = 4

⇒ A(y) = 5

A(4 ,5)

⇒ B(x) = 8

⇒ B(y) = 7

B(8,7)

⇒ C(x) = 14

⇒ C(y) = 5

C(14,5)

Let the fourth vertex of the smaller quadrilateral be (a,b)

⇒ a = 8

⇒ b = 3

Three vertices of larger quadrilateral are A(4 ,5) , B(8,7) & C(14,5) and the fourth vertex of the smaller quadrilateral is (8,3)

The three vertices are (3, 5), (9, 13), (10, 6)

Length of first side = √((9 - 3)² + (13 - 5)²)

⇒ Length of first side = √(6² + 8²)

⇒ Length of first side = 10 units

Length of second side = √((10 - 9)² + (6 - 13)²)

⇒ Length of second side = √(1² + 7²)

⇒ Length of second side = √50 units

Length of third side = √((10 - 3)² + (6 - 5)²)

⇒ Length of third side = √(7² + 1²)

⇒ Length of third side = √50 units

Since the length of second and third sides are equal so the triangle is isosceles

Length of height

Length of height = √75 units

Area = 0.5 × base × height

⇒ Area = 12.5 × 1.732

⇒ Area = 12.5 × 1.732 = 10.768 sq. units

The three coordinates are ( - 1, 5) , (3, 7) , (3, 1)

Let the centroid be C(x₁, y₁)

The Coordinates of centroid is

Centre of circle is (1, 2)

Point (3, 2)

Let the coordinates of the other end be (x₁,y₁)

Using the section formula for mid points

⇒ 3 + x₁ = 2

⇒ x₁ = - 1

⇒ 2 + y₁ = 4

⇒ y₁ = 2

Answer : The Other point is ( - 1 , 2)

The three points are (1, 8) ,(2, 5) ,(3, 7)

Slope of First two points

Slope of First two points = - 3

The Equation of line between first two points :

y - 8 = - 3(x - 1)

⇒ y - 8 = - 3x + 3

⇒ 3x + y = 11

We put the point (3, 7) in the above equation.

On solving we will find the point doesn’t satisfy the equation.

Hence the three points doesn’t lie on a straight line.

The two points are ( - 1, 4) ,(1,2)

Slope of two points

Slope of two points = - 1

The Equation of line between two points :

y - 2 = - 1 (x–1)

⇒ x + y - 3 = 0

Putting x = 0 in the above equation we get y = 3

Putting y = 0 in the above equation we get x = 3

So the other two points are (0,3) (3,0)

Let the common difference between the x - coordinate be d_x , between the y - coordinate be d_y

The three points can be written in terms of their common difference as (x₁, y₁), (x₁ + d_x, y₁ + d_y), (x₁ + 2d_x, y₁ + 2d_y)

Slope of first two points

⇒ Slope of first two points

Slope of last two points

⇒ Slope of last two points

Since the slope of first two points and last two points are same hence they are on the same line.

Hence proved

The three points are say A(x₁, y₁), B(x₂, y₂), C(x₃, y₃)

Since they lie on a line so slope of any two points are always equal

…Equation (i)

The other set of three points are say P(3x₁ + 2y₁,3x₁ – 2y₁), Q(3x₂ + 2y₂,3x₂ – 2y₂), R(3x₃ + 2y₃,3x₃ – 2y₃)

Since they also lie on a line so slope between any two points is always equal

Applying Componendo and dividendo we get

Applying Invertendo we get

…Equation (ii)

Since Equation (i) & Equation (ii) are similar so the points P,Q and R lie on the line joining A,B & C

Hence Proved

Yes it is possible if we take multiples of 2 and 3

Slope of the line

⇒ Slope of the line = 2

Equation of line :

y - 2 = 2(x - 1)

⇒ 2x - y = 0

⇒ y = 2x

Value of y for consecutive natural numbers :

Slope of the line

⇒ Slope of the line

Equation of line :

⇒ 3(y - 5) = 2(x - 2)

⇒ 2x - 3y + 11 = 0 …Equation (i)

Putting x = x + 3 and y = y + 2 in the above equation we get

2(x + 3) - 3(y + 2) + 11 = 0

⇒ 2x - 3y + 11 = 0

Since it again gives the same equation as (i) so it is a point on this line.

Hence Proved

Slope of the line

⇒ Slope of the line = 2

Equation of line :

y - 2 = 2(x - 7)

⇒ 2x - y - 12 = 0

Putting x = x and y = 2x + 3 in the above equation we get

2x - (2x + 3) - 12 = 0

which satisfies the above equation

Hence Proved

(i) Let the y coordinate be a

tan 60⁰

⇒ a = 2√3

The y coordinate is 2√3

(ii) Slope = tan 60⁰

⇒ Slope of line = √3

(iii) Equation of line:

y - 0 = √3(x - 1)

⇒ √3x - y - √3 = 0

Slope of diagonal AC

⇒ Slope of diagonal AC = 1

Since diagonals of a square are perpendicular to each other

Slope of diagonal AC × Slope of diagonal BD = - 1

⇒ Slope of diagonal BD = - 1

Origin (0,0) is a point on the diagonal

Equation of diagonal BD:

y = - x

⇒ x + y = 0

Hence Proved

Length of x intercept (a) = 3 units

Length of y intercept (b) = 3 units

According to slope intercept form , the equation of the line is

⇒ x + y = 3

Proved

Radius = 5 units

Equation of circle:

x² + y² = 25

Consider x = 0

⇒ y² = 25

⇒ y = ±5

Again Consider y = 0

⇒ x² = 25

⇒ x = ±5

Consider x = y

2x² = 25

Consider x = - y

2x² = 25

Hence the Eight points are:

Two diametrically Opposite points are (0,1),(2,3)

Equation of circle for two Diametrically Opposite points:

(x - 0)(x - 2) + (y - 1)(y - 3) = 0

⇒ x² - 2x + y² - 4y + 3 = 0

When the circle cuts the x axis the y coordinate is 0

⇒ x² - 2x + 3 = 0

Discriminant = ( - 2)² - 4 × 1 × 3

Discriminant = - 8

Since Discriminant is negative so the roots are imaginary

Hence the circle doesn’t cut the x - axis at any point.

The X and Y axis meets the circle at point O

The line joining the two points (0,2) and (4,0) forms the diameter of the circle.

Let the centre of the circle be C(x, y)

Using Section Formula for mid points

⇒ C(x) = 2

⇒ C(y) = 1

The coordinate of Center is (2, 1)

Radius of Circle = √((4 - 2)² + (0 - 1)²)

⇒ Radius = √5 units

So the equation of circle is:

(x - 2)² + (y - 1)² = 5

⇒ x² + y² - 4x - 2y + 5 = 5

x² + y² - 4x - 2y = 0