Question

If $p_1,p_2$ are the perpendicular distances from the origin to the two straight lines which are perpendicular to each other, then the locus of the point of intersection of the perpendicular lines is

• A. $x^2+y^2=p_1+p_2$
• B. $x^2+y^2=P_1^2+P_2^2$
• C. $\frac{x_1^2}{p_1^2}+\frac{x_2^2}{p_2^2}=1$
• D. $xy=p_1{p}_2$

Answer 1

Answer: (B)

$x^2+y^2=P_1^2+P_2^2$

let two perpendicular line is

$ax+by+c=0–\left(i\right)$

$bx-ay+d=0–\left(ii\right)$

$P_1=\frac{c}{\sqrt{a^2+b^2}}$

$P_1=\frac{d}{\sqrt{a^2+b^2}}$

Solving $\left(i\right)$ & $\left(ii\right)$ $x=\frac{-(ac+bd)}{a^2+b^2},y=\frac{ad-bc}{a^2+b^2}$

$x^2+y^2=\frac{(ac+bd)}{(a^2+b^2)}+\frac{(ad-bc{)}^2}{(a^2+b^2{)}^2}$

$\Rightarrow x^2+y^2=\frac{a^2{c}^2+b^2{d}^2+2abcd+a^2{d}^2+b^2{c}^2-2abcd}{(a^2+b^2{)}^2}$

$\Rightarrow x^2+y^2=\frac{a^2(c^2+d^2)+b^2(c^2+d^2)}{(a^2+b^2{)}^2}$

$\Rightarrow x^2+y^2=\frac{(a^2+b^2)(c^2+d^2)}{(a^2+b^2{)}^2}$

$\Rightarrow x^2+y^2=\frac{c^2+d^2}{a^2+b^2}$

$\Rightarrow x^2+y^2=\frac{c^2}{a^2+b^2}+\frac{d^2}{a^2+b^2}$

$\Rightarrow x^2+y^2=p_1^2+p_2^2$