Question

If the distance of a point $(x_1,y_1)$ from each of the two straight lines, which pass through the origin of coordinates, is $d$, then the two lines are given by

• A. ${(x{y}_1–y{x}_1)}^2=d^2(x^2+y^2)$
• B. ${(x_1{y}_1–yx)}^2=(x^2+y^2)$
• C. ${(x{y}_1+y{x}_1)}^2=(x^2-y^2)$
• D. $\left(x^2-y^2\right)=2\left(x_1+y_1\right)$

Answer 1

The correct option is A

${(x{y}_1–y{x}_1)}^2=d^2(x^2+y^2)$

Explanation for correct option

Any line passing through the origin is of the form $y=mx$ or $y-mx=0$

We know that the distance of the point $\left(p,q\right)$ from the line $ax+by+c=0$ is $d=\frac{\left|ap+bq+c\right|}{\sqrt{a^2+b^2}}$

Therefore the distance of the point $(x_1,y_1)$ from the line $y-mx=0$ is $d=\frac{\left|-m{x}_1+y_1\right|}{\sqrt{{\left(-m\right)}^2+1^2}}$

$$\begin{gathered} \therefore d=\frac{\left|-m{x}_1+y_1\right|}{\sqrt{{\left(-m\right)}^2+1^2}} \\ \Rightarrow {\left(d\sqrt{{\left(-m\right)}^2+1^2}\right)}^2={\left(\left|-m{x}_1+y_1\right|\right)}^2 \\ \Rightarrow d^2\left(m^2+1\right)={\left(y_1-m{x}_1\right)}^2 \\ \Rightarrow d^2\left({\left(\frac{y}{x}\right)}^2+1\right)={\left(y_1-\left(\frac{y}{x}\right)x_1\right)}^2\left[\because y=mx\right] \\ \Rightarrow d^2\left(x^2+y^2\right)=x^2{\left(\frac{y_{1}x-y{x}_1}{x}\right)}^2 \\ \Rightarrow d^2\left(x^2+y^2\right)={\left(y_{1}x-y{x}_1\right)}^2 \end{gathered}$$

Hence, option (A) is correct i.e. ${(x{y}_1–y{x}_1)}^2=d^2(x^2+y^2)$