question_answer

Let coordinates of the points \[A\] and \[B\] are (5, 0) and (0, 7) respectively. P and Q are the variable points lying on the x-and y-axis respectively so that PQ is always perpendicular to the line AB. Then locus of the point of intersection of BP and AQ is

A) \[{{x}^{2}}+{{y}^{2}}-5x+7y=0\]

B) \[{{x}^{2}}+{{y}^{2}}+5x-7y=0\]

C) \[{{x}^{2}}+{{y}^{2}}+5x+7y=0\]

D) \[{{x}^{2}}+{{y}^{2}}-5x-7y=0\]

Correct Answer: D

Solution :

Let the equation of PQ be \[\frac{x}{a}+\frac{y}{b}=1\]  ...(i)
Then equation of BP is \[\frac{x}{a}+\frac{y}{7}=1\]	...(ii)
and equation of AQ is \[\frac{x}{5}+\frac{y}{b}=1\]	...(iii)
Since \[AB\bot PQ,\] therefore, \[-\frac{b}{a}\times -\frac{7}{5}=-1\]     \[\Rightarrow \]\[5a+7b=0\]		...(iv)

Now, slope of \[BP\,\operatorname{is}-\frac{7}{a}\]and slope of \[AQ\]is \[-\frac{b}{5}\]

So, product of these slopes is \[\frac{7b}{5a}=-1\]        [from (iv)]

\[\therefore BP\,\operatorname{and}\,AQ\].intersect each other at right angle. So, the locus of the point of intersection of \[BP\]and \[AQ\] is a circle with \[AB\] as diameter. Thus equation of locus is

\[x\left( x-5 \right)+y\left( y-7 \right)=0\] \[\Rightarrow {{x}^{2}}+{{y}^{2}}-5x-7y=0\]