WBJEE · Maths · Straight Lines
A line cuts the \(X\) -axis at \(A(5,0)\) and the \(Y\) -axis at \(B(0,-3)\). A variable line \(P Q\) is drawn perpendicular to \(A B\) cutting the \(X\) -axis at \(P\) and the \(Y\) -axis at \(Q\). If \(A Q\) and \(B P\) meet at \(R\), then the locus of \(R\) is
- A \(x^{2}+y^{2}-5 x+3 y=0\)
- B \(x^{2}+y^{2}+5 x+3 y=0\)
- C \(x^{2}+y^{2}+5 x-3 y=0\)
- D \(x^{2}+y^{2}-5 x-3 y=0\)
Answer & Solution
Correct Answer
(A) \(x^{2}+y^{2}-5 x+3 y=0\)
Step-by-step Solution
Detailed explanation
Equation of line \(A B\) is \[ \begin{array}{l} \frac{x}{5}+\frac{y}{-3}=1 \\ 3 x-5 y=15 \end{array} \] Perpendicular line to \(A B\) is \[ 5 x+3 y=\lambda \] Coordinate of \(P\) is \(\left(\frac{\lambda}{5}, 0\right)\) and coordinate of \(Q\) is \((0, \lambda / 3)\) Now,…
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