Question

If the midpoints of sides of the triangle pqr are (1,2) (0,1) (1,0), then find the coordinates of the vertices of the triangle.

Answer 1

refer to the attachment

Answer 2

Answer:

Coordinates of P( 0 , 3 ) , Q( 2 , 1 ) and R( 0 , -1 ).

Step-by-step explanation:

Given: Mid Point of sides of Triangle PQR are L( 1 , 2 ) , M( 0 , 1 ) and N( 1 , 0 )

To find: Coordinates of Vertices of triangle.

Let say the coordinates of vertices of ΔPQR are P( x , y ) , Q( p , q ) & R( r , s )

We use Mid point formula to form equations then we find value of all variables

Coordinates of Mid Point of two points = $(\frac{x_1+x_2}{2}\:,\:\frac{y_1+y_2}{2})$

Using points P, Q & L

Coordinates of L = $(\frac{x+p}{2}\:,\:\frac{y+q}{2})$

$(\,1\:,\:2\,)\:=\:(\frac{x+p}{2}\:,\:\frac{y+q}{2})$

⇒ $\frac{x+p}{2}=1$ ⇒ x + p = 2 ..................(1)

⇒ $\frac{y+q}{2}=2$ ⇒ y + q = 4 .................(2)

Using points Q, R & N

Coordinates of N = $(\frac{p+r}{2}\:,\:\frac{q+s}{2})$

$(\,1\:,\:0\,)\:=\:(\frac{p+r}{2}\:,\:\frac{q+s}{2})$

⇒ $\frac{p+r}{2}=1$ ⇒ p + r = 2 ..................(3)

⇒ $\frac{q+s}{2}=0$ ⇒ q + s = 0 .................(4)

Using points P, R & M

Coordinates of M = $(\frac{x+r}{2}\:,\:\frac{y+s}{2})$

$(\,0\:,\:1\,)\:=\:(\frac{x+r}{2}\:,\:\frac{y+s}{2})$

⇒ $\frac{x+r}{2}=0$ ⇒ x + r = 0 ..................(5)

⇒ $\frac{y+s}{2}=1$ ⇒ y + s = 2 .................(6)

Now, Subtract (5) from (1), we get

p - r = 2  ⇒ p = 2 + r

put value of p in (3), we get

2 + r + r = 2 ⇒ 2r = 0 ⇒ r = 0

⇒ p = 2

From (1), we get

⇒ x = 0

Now, Subtract (6) from (2), we get

q - s = 2  ⇒ q = 2 + s

put value of p in (4), we get

2 + s + s = 0 ⇒ 2s = -2 ⇒ s = -1

⇒ q = 1

From (2), we get

⇒ y  = 3

Therefore, Coordinates of P( 0 , 3 ) , Q( 2 , 1 ) and R( 0 , -1 ).