Question

P (x, y), Q (–2, – 3) and R (2, 3) are the vertices of a right triangle PQR right angled at P. Find the relationship between x and y. Hence, find all possible values of x for which y = 2.

Answer 1

We know

\(PQ^2 + PR^2 = QR^2\)    (Pythagoras theorem)

\((x+2)^2 + (y+3)^2 + (x-2)^2+(y-3)^2=(-2-2)^2 + (-3-3)^2\)

\(x^2 +4+4x+y^2+9+6y+x^2+4-4x+y^2+9-6y=16+36\)

\(2(x^2 +y^2) +2(4+9) +52\)

\(x^2+y^2+13=26\)

\(x^2+y^2=13\)  (Required relation)

For y = 2, \(x^2=13-2^2=9\)

\(x\pm3\)

\(\therefore\) Possible values of x when y = 2 are –3, 3.