Question

P(2,2),Q(3,1) and R(4,2) are midpoint of sides of triangle ABC, find vertices of triangle ABC.​

Answer 1

$\huge{\green{\underline{\underline{Answer:-}}}}$

Given:

Triangle ABC,

P, Q and R are midpoint of side AB, BC and CD

To find:

Coordinates of vertices A, B and C = ?

Solution:

Note: If A (x, y) and B ( a, b) are two points then coordinates of midpoint of seg AB are

$\\ \\ \sf {\bigg(\dfrac{x + a}{2} , \: \dfrac{y + b}{2} \bigg) }$

A, B and C are vertices of the triangle

Let x and y coordinates of A, B and C be

A = x, a

B = a, b

C = p , q

Given: P (2, 2)

x coordinate = 2

x coordinate = 2y coordinate = 2

Q ( 3, 1 )

x coordinate = 3

x coordinate = 3y coordinate = 1

R ( 4, 2 )

x coordinate = 4

x coordinate = 4y coordinate = 2

P is the midpoint of AB

Q is midpoint of BC

R is midpoint of AC

$\\ \\ x \: \: coordinate \: \: of \: \: P = \dfrac{x + a}{2} \\ \\ x + a = 4 - - (1)\\ \\ y \: \: coordinate \: \: of \: \: P = \dfrac{y + b}{2} \\ \\ y + b = 4 - - (2)$

$\\ \\ x \: \: coordinate \: \: of \: \: Q= \dfrac{a + p}{2} \\ \\ a + p = 6 - - (3) \\ \\ y\: \: coordinate \: \: of \: \: Q= \dfrac{b + q}{2} \\ \\ b + q = 2 - - (4)$

$\\ \\ x \: \: coordinate \: \: of \: \: R= \dfrac{x + p}{2} \\ \\ x + p = 8 - - - (5)\\ \\ y\: \: coordinate \: \: of \: \: R= \dfrac{y + q}{2} \\ \\ y + q = 4 - - (6)$

a + p = 6 ---( from 6 )

a = 6 - p

Substituting value of a in equation (1)

x + a = 4

x + 6 - p = 4

x - p = -2

x + p = 8 -- from (5)

p = 8 - x

Substituting p = 8 - x in equation x - p = -2

x - (8 - x) = -2

x - 8 + x = -2

x + x = 8 - 2

2x = 6

x = 3

x + p = 8 -- from (5)

3 + p = 8 -- ( x = 3)

p = 8 - 3

p = 5

x + a = 4 --- from ( 1)

3 + a = 4 -- ( x = 3)

a = 4 - 3

a = 1

y + b = 4 -- from ( 2 )

y = 4 - b

Substituting y = 4 - b in equation (6)

y + q = 4

4 - b + q = 4

q - b = 4 -4

q - b = 0

q = b

b + q = 2 -- from (4)

b + b = 2 --- ( q = b)

2b = 2

b = 1

Then, q = 1 -- ( b = q)

y + b = 4 --- ( from 2)

y + 1 = 4 -- ( b = 1)

y = 4 - 1

y = 3

Answer:

Coordinates of vertices of triangle ABC

• Coordinates of A ( x, y) = (3, 3)
• Coordinates of B (a, b) = (1, 1)
• Coordinates of C (p,q) = (5, 1)

Verification:

Verify by substituting the value, if we get proper answer our answer is correct

For P, x and y coordinate = 2,2

x coordinate = x+a/2 = 3 + 1/ 2 = 2

y coordinate = y + b/2 = 3 + 1/2 = 2

Hence proved

For Q, x and y coordinate = 3,1

x coordinate = a + p/ 2 = 1+5/2 = 3

y coordinate = b + q / 2 = 1 + 1/2 = 1

Hence proved

For R, x and y coordinate = 4,2

x coordinate = x + p/ 2 = 3 + 5/2 = 4

y coordinate = y + q / 2 = 3 + 1/2 = 2

Hence proved

Knowledge booster:

• Try to take same variables together for substitution and solving as for easy way.
• Remember formulas and choose suitable variables
• Solve more such questions to get good hold on it