Question
A line passes through $(1,0)$. The slope of the line,for which its intercept between $y = x - 2$ and $y = - x + 2$ subtends a right angle at the origin, is :
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Solution
Sol. [Key Concept: The intercept of a line through $(1,0)$ subtends a right angle at $O$ ⇒ a slope condition from $m_{1}m_{2 = - 1}$.]
Let the line be $y = m(x - 1)$
Equation of pair of lines is $(y - x + 2)(y + x - 2) = 0$
$$\Rightarrow y^{2} - x^{2} + 4x - 4 = 0 $$Equation of pair $OA$ and $OB$ is obtained by homogenisation given by
$${y^{2} - x^{2} + 4x\left( \frac{mx - y}{m} \right) - 4\left( \frac{mx - y}{m} \right)^{2} = 0 }{\because\ OA\bot OB }$$∴ Coefficient of $x^{2} +$ Coefficient of $y^{2} = 0$
$$\Rightarrow 1 - 1 + 4 - 4 + \frac{4}{m^{2}} = 0 \Rightarrow m \rightarrow \infty $$$\therefore\ $ Line is given by $x = 1$.
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