Question

Let $PQR$ be a right angled isosceles triangle, right angled at $P (2,1)$. If the equation of the line $QR$ is $2x + y = 3$, then the equation representing the pair of lines PQ and PR is

• A. $3x^2 - 3 y^2 + 8 xy + 20 x + 10 y + 25 = 0$
• B. $3x^2 - 3 y^2 + 8 xy - 20 x - 10 y + 25 = 0$
• C. $3x^2 - 3 y^2 + 8 xy + 10 x + 15 y + 25 = 0$
• D. $3x^2 - 3 y^2 - 8 xy + 10 x + 15 y + 25 = 0$

Answer 1

Answer: (B)

$3x^2 - 3 y^2 + 8 xy - 20 x - 10 y + 25 = 0$

Answer 2

Let S be the mid-point of QR and given $\triangle$ PQR is an isosceles.
Therefore, PS $\perp$ QR and S is mid-point of hypotenuse,
$\therefore$ S is equidistant from P, Q, R.
$\therefore PS = QS = RS$
SInce, $\angle P = 90^\circ$
and $\angle Q = \angle R$
But $\angle P + \angle Q + \angle R = 180^\circ$
$\therefore 90^\circ + \angle Q + \angle R = 180^\circ$
$\Rightarrow$ $\angle Q = \angle R = 45^\circ$
Now, slope of QR is - 2 .
But $QR \perp PS$
$\therefore$ Slope of PS is 1/2.
Let m be the slope of P
$\therefore \tan ( \pm 45^\circ) = \frac{ m - 1/ 2 }{ 1 - m \, ( - 1 / 2) }$
$\Rightarrow \pm 1 \frac{ 2m - 1}{ 2 + m}$
$\Rightarrow m = 3, - 1 / 3$
$\therefore$ Equations of PQ and PR are $y - 1 = 3 (x - 2 )$ and $y - 1 = - \frac{1}{3} \, (x - 2)$
or 3 ( y - 1) + (x - 2) = 0
Therefore, joint equation of PQ and PR is
[ 3 (x - 2) - ( y - 1) ] [ (x - 2) + 3 ( y - 1) ] = 0
$\Rightarrow 3 ( x - 2 )^2 - 3 ( y - 1)^2 + 8 (x - 2) \, ( y - 1) = 0$
$\Rightarrow 3x^2 - 3y^2 + 8 xy - 20 x - 10 y + 25 = 0$