In an isosceles Δ PQR, PQ = PR. The base QR lies on the x–axis, P lies on the y– axis and 2x – 3y + 9 = 0 is the equation of PQ. Find the equation of the straight line along PR.

Here it’s given that in an isosceles PQR, PQ = PR. The base QR lies on the x–axis, P lies on the y– axis and 2x – 3y + 9 = 0 is the equation of PQ.

The point P lies on the y–axis, so we have to put x = 0 in the equation for PQ

i.e.2x –3y + 9 = 0

⇒ 2(0)–3y + 9 = 0

⇒ –3y + 9 = 0

⇒ –3y = –9

⇒

Therefore, the point P is (0,3).

Now, to find the point Q which is on the x axis ,we have to substitute 0 in the place of y in the given equation QR.

2x –3y + 9 = 0

⇒ 2x–3(0) + 9 = 0

⇒ 2x + 9 = 0

⇒ 2x = –9

⇒

Therefore,the point Q becomes (

Now to find the equation of PR, where P is (0,3) and Q is (,we

have to use ,

⇒

⇒

⇒

⇒ –9y + 27 = 2(–3x)

⇒ –9y + 27 = –6x

⇒ –9y + 27 + 6x = 0

⇒ 2x–3y + 9 = 0(Divide by 3 on both the sides)