Question

Line L has intercepts a and b on the coordinate axes. when the axes are rotated through a given angle, keeping the origin fixed, the same line L has intercepts p and q. Then,

• A. $a^2+b^2=p^2+q^2$
• B. $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}+\frac{1}{q^2}$
• C. $a^2+p^2=b^2+q^2$
• D. $\frac{1}{a^2}+\frac{1}{p^2}=\frac{1}{b^2}+\frac{1}{q^2}$

Answer 1

The correct option is B $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}+\frac{1}{q^2}$


As L has intercepts and b on the axes, the equation of L is $\frac{x}{a}+\frac{y}{b}=1$
Let the x-axis and the y-axis be rotated through an angle $\theta$ in the anticlockwise direction. In the new system, the intercepts are p and q (OP= p,OQ=q). Therefore , the equation of L w.r.t.
new coordinate system becomes
$\frac{x}{p}+\frac{y}{q}=1$
As the origin is fixed in rotation, the distance of line from the origin in both the cases should be same. Hence, we get
$d=\left|\begin{array}{c}-\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}\end{array}\right|=\left|\begin{array}{c}\frac{1}{\sqrt{\frac{1}{p^2}+\frac{1}{q^2}}}\end{array}\right|\text{(Using the formula for distance of a point from a line)}or\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}+\frac{1}{q^2}$