Straight Line and Pair of Straight Lines 1 Question 59
59. $A$ line cuts the $X$-axis at $A(7,0)$ and the $Y$-axis at $B(0,-5)$. $A$ variable line $P Q$ is drawn perpendicular to $A B$ cutting the $X$-axis in $P$ and the $Y$-axis in $Q$. If $A Q$ and $B P$ inters at $R$, find the locus of $R$.
(1990, 4M)
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Solution:
- The equation of the line $A B$ is
$$ \frac{x}{7}+\frac{y}{-5}=1 $$
$$ \Rightarrow \quad 5 x-7 y=35 $$
Equation of line perpendicular to $A B$ is
$$ 7 x+5 y=\lambda $$
It meets $X$-axis at $P(\lambda / 7,0)$ and $Y$-axis at $Q(0, \lambda / 5)$.
The equations of lines $A Q$ and $B P$ are $\frac{x}{7}+\frac{5 y}{\lambda}=1$ and $\frac{7 x}{\lambda}-\frac{y}{5}=1$, respectively.
Let $R(h, k)$ be their point of intersection of lines $A Q$ and $B P$.
Then, $\quad \frac{h}{7}+\frac{5 k}{\lambda}=1$
and $\quad \frac{7 h}{\lambda}-\frac{k}{5}=1$
$\Rightarrow \quad \frac{1}{5 k} 1-\frac{h}{7}=\frac{1}{7 h} 1+\frac{k}{5} \quad$ [on eliminating $\lambda$ ]
$\Rightarrow \quad h(7-h)=k(5+k)$
$\Rightarrow \quad h^{2}+k^{2}-7 h+5 k=0$
Hence, the locus of a point is
$$ x^{2}+y^{2}-7 x+5 y=0 \text {. } $$
