Question

A line cuts the x-axis at \[A\left( 7,0 \right)\] and the y-axis at \[B\left( 0,-5 \right)\] . A variable line \[PQ\] is drawn perpendicular to \[AB~\]cutting the x-axis at $P$ and the y-axis at in $Q$ . If \[AQ\] and $BP$ intersect at $R$ , the locus of $R$ is
A. \[{{x}^{2}}+{{y}^{2}}+7x-5y=0\]
B. \[{{x}^{2}}+{{y}^{2}}-7x+5y=0\]
C. \[5x-7y=35\]
D. None of these

Answer 1

Hint: We have to find the locus of $R$ . First find the equation of line \[AB~\]. Since \[PQ\] is drawn perpendicular to \[AB~\] , the product of their slopes is equal to $1$ , that is, \[{{m}_{AB}}\times {{m}_{PQ}}=-1\] . Then find the equation of line joining \[AQ\] and then \[BP\] . \[AQ\] and $BP$ intersect at $R(h,k)$ . Then the equation of line joining \[AQ\] and the line \[BP\] is written in terms of $(h,k)$ . By suitable substitution, we will get the locus of $R$.

Complete step by step answer:
Given that a line cuts the x-axis at \[A\left( 7,0 \right)\] and the y-axis at \[B\left( 0,-5 \right)\] . \[PQ\] is drawn perpendicular to \[AB~\]cutting the x-axis at $P$ and the y-axis at in $Q$ . Also given that \[AQ\] and $BP$ intersect at $R$. We have to find the locus of $R$ .

The equation of a line passing through the points $(a,0)$ and $(0,b)$ or making intercepts $a$ and $b$ on the x-and y-axis respectively is given by
$\dfrac{x}{a}+\dfrac{y}{b}=1$
Here, the points are \[A\left( 7,0 \right)\] and \[B\left( 0,-5 \right)\] . Therefore, the equation of line \[AB~\] is
$\dfrac{x}{7}+\dfrac{y}{-5}=1...(i)$
\[PQ\] is drawn perpendicular to \[AB~\]cutting the x-axis at $P(\alpha ,0)$ and the y-axis at in $Q(0,\beta )$ .
Therefore, $\text{slope of AB}\times \text{slope of PQ}=-1$
Slope of \[AB~={{m}_{AB}}=\dfrac{-5-0}{0-7}=\dfrac{5}{7}\]
Slope of \[PQ~={{m}_{PQ}}=\dfrac{\beta -0}{0-\alpha }=-\dfrac{\beta }{\alpha }\]
Hence, \[{{m}_{AB}}\times {{m}_{PQ}}=-1\]
Substituting the values, we get
$\dfrac{5}{7}\times -\dfrac{\beta }{\alpha }=-1$
From this, we get
$\dfrac{\beta }{\alpha }=\dfrac{7}{5}...(i)$
Now, the equation of line joining \[A\left( 7,0 \right)\] and $Q(0,\beta )$ , that is, \[AQ\] is
$\dfrac{x}{7}+\dfrac{y}{\beta }=1...(ii)$
The equation of line joining \[B\left( 0,-5 \right)\] and $P(\alpha ,0)$ , that is, \[BP\] is
$\dfrac{x}{\alpha }+\dfrac{y}{-5}=1...(iii)$
\[AQ\] and $BP$ intersect at $R(h,k)$ . So the equation $(ii)$ can be written as
$\dfrac{h}{7}+\dfrac{k}{\beta }=1$
Rearranging the terms, we will get
$\dfrac{k}{\beta }=1-\dfrac{h}{7}$
$\Rightarrow \dfrac{k}{\beta }=\dfrac{7-h}{7}$
$\Rightarrow \beta =\dfrac{7k}{7-h}...(iv)$
Similarly, equation $(iii)$ can be written as
$\dfrac{h}{\alpha }+\dfrac{k}{-5}=1$
Rearranging the terms, we will get
$\dfrac{h}{\alpha }=1+\dfrac{k}{5}$
$\Rightarrow \dfrac{h}{\alpha }=\dfrac{5+k}{5}$
$\Rightarrow \alpha =\dfrac{5h}{5+k}...(v)$
Now, substitute $(iv)\text{ and }(v)\text{ in }(i)$, we will get
$\dfrac{\dfrac{7k}{7-h}.}{\dfrac{5h}{5+k}}=\dfrac{7}{5}$
Solving this, we get
$\dfrac{7k(5+k)}{(7-h)5h}=\dfrac{7}{5}$
Simplifying further gives,
$\dfrac{k(5+k)}{(7-h)h}=1$
$\Rightarrow \dfrac{5k+{{k}^{2}}}{7h-{{h}^{2}}}=1$
$\Rightarrow 5k+{{k}^{2}}=7h-{{h}^{2}}$
$\Rightarrow 5k+{{k}^{2}}-7h+{{h}^{2}}$=0
Substituting $h=x\text{ and }k=y$ and rearranging, we get
${{x}^{2}}+{{y}^{2}}-7x+5y$=0

So, the correct answer is “Option B”.

Note: The equation of a line passing through the points $(a,0)$ and $(0,b)$ or making intercepts $a$ and $b$ on the x-and y-axis respectively is given by $\dfrac{x}{a}+\dfrac{y}{b}=1$ . This equation is used here many times. Also $\text{slope of AB}\times \text{slope of PQ}=-1$ is another important equation.