Question

A line passes through (3, 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 may be

• A. $\sqrt{2}$
• B. $-\sqrt{2}$
• C. $\frac{1}{\sqrt{3}}$
• D. $-\frac{1}{\sqrt{2}}$

Answer 1

Answer: (D)

$-\frac{1}{\sqrt{2}}$

Given line

$y=x-2\Rightarrow x-y-2=0----\left(1\right)$

$y=-x+2\Rightarrow x+y-2=0----\left(2\right)$

On multiplying eq (1) and (2)

$(x-y-2)(x+y-2)=0$

$x^2+4-4x-y^2=0$

$x^2-y^2-4x+4=0---\left(3\right)$

Equation of line from point $(3,0)$ with slope m

$y=mx-3m$

$1=\frac{mx-y}{3m}$

From eq (3)

$x^2-y^2-4x\left(\frac{mx-y}{3m}\right)+4{\left(\frac{mx-y}{3m}\right)}^2=0$

$x^2-y^2-\left(\frac{4m{x}^2-4xy}{3m}\right)+4\left(\frac{m^2{x}^2+y^2-2mxy}{9m^2}\right)=0$

$9m^2{x}^2-9m^2{y}^2-12m^2{x}^2+12mxy+4m^2{x}^2+4y^2-8mxy=0$

$\left(m^2\right)x^2+(-9m^2+4)y^2+4mxy=0$

Since line subtends right angle

$m^2-9m^2+4=0$

$8m^2=4$

$m^2=\frac{1}{2}$

$m=\pm \frac{1}{\sqrt{2}}$