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

• A. $\pm 2/3$
• B. $\pm 3/2$
• C. $1$
• D. none of these

Answer 1

The correct option is D none of these

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 $(1,0)$ with slope m

$y=mx-m$

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

From eq (3)

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

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

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

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

Since line subtends right angle

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

Hence no solution