Question

A line cuts the \(x\)-axis at \(A(7,0)\) and the \(y\)-axis at \(B(0,-5)\). A variable line \(PQ\) is drawn perpendicular to \(AB\) cutting the \(x\)-axis and \(y\)-axis at \(P\) and \(Q\) respectively. If \(AQ\) and \(BP\) intersect at \(R\), then the locus of \(R\) is:

• A. \(x^2+y^2+7x-5y=0\)
• B. \(x^2+y^2-7x+5y=0\)
• C. \(x^2+y^2-3x+4y=0\)
• D. \(x^2+y^2+6x+7y=0\)

Hint:

In locus problems:
Assign variables to moving intercepts.
Use slope conditions for perpendicular lines.
Eliminate parameters to obtain the locus equation. Such loci often turn out to be circles.

Answer 1

The Correct Option is B

Step 1: Find the equation and slope of line \(AB\). Line \(AB\) cuts the axes at: \[ A(7,0), \quad B(0,-5) \] Slope of \(AB\): \[ m_{AB}=\frac{-5-0}{0-7}=\frac{5}{7} \]
Step 2: Write the equation of variable line \(PQ\). Since \(PQ \perp AB\), its slope is: \[ m_{PQ}=-\frac{7}{5} \] Let \(P(a,0)\) and \(Q(0,b)\). Slope of \(PQ\): \[ \frac{b-0}{0-a}=-\frac{b}{a} \] Equating slopes: \[ -\frac{b}{a}=-\frac{7}{5} \Rightarrow \frac{b}{a}=\frac{7}{5} \Rightarrow b=\frac{7a}{5} \]
Step 3: Find equations of lines \(AQ\) and \(BP\). Equation of \(AQ\) through \(A(7,0)\) and \(Q(0,b)\): \[ \frac{y-0}{x-7}=\frac{b-0}{0-7} \Rightarrow y=-\frac{b}{7}(x-7) \] Equation of \(BP\) through \(B(0,-5)\) and \(P(a,0)\): \[ \frac{y+5}{x}=\frac{0+5}{a} \Rightarrow ay=5x-5a \]
Step 4: Find coordinates of intersection point \(R(x,y)\). Substitute \(b=\frac{7a}{5}\) into equation of \(AQ\): \[ y=-\frac{a}{5}(x-7) \] From \(BP\): \[ y=\frac{5x-5a}{a} \] Equating both expressions for \(y\) and simplifying, we eliminate \(a\) to obtain: \[ x^2+y^2-7x+5y=0 \]
Hence, the locus of point \(R\) is \[ \boxed{x^2+y^2-7x+5y=0} \]