A line cuts the X-axis at A(5,0) and the Y-ax | vedclass

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(A) The equation of line $AB$ is $\frac{x}{5} + \frac{y}{-3} = 1$, which simplifies to $3x - 5y = 15$.Since line $PQ$ is perpendicular to $AB$, its eq

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$x^{2}+y^{2}-5x+3y=0$

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(A) The equation of line $AB$ is $\frac{x}{5} + \frac{y}{-3} = 1$, which simplifies to $3x - 5y = 15$.Since line $PQ$ is perpendicular to $AB$, its equation is of the form $5x + 3y = \lambda$.The coordinates of $P$ (where $y=0$) are $(\frac{\lambda}{5}, 0)$ and the coordinates of $Q$ (where $x=0$) are $(0, \frac{\lambda}{3})$.The equation of line $AQ$ passing through $A(5,0)$ and $Q(0, \frac{\lambda}{3})$ is $\frac{x}{5} + \frac{y}{\lambda/3} = 1$, which gives $\frac{x}{5} + \frac{3y}{\lambda} = 1$. Thus, $\frac{1}{\lambda} = \frac{1}{3y}(1 - \frac{x}{5})$.The equation of line $BP$ passing through $B(0,-3)$ and $P(\frac{\lambda}{5}, 0)$ is $\frac{x}{\lambda/5} + \frac{y}{-3} = 1$, which gives $\frac{5x}{\lambda} - \frac{y}{3} = 1$. Thus, $\frac{1}{\lambda} = \frac{1}{5x}(\frac{y}{3} + 1)$.Equating the two expressions for $\frac{1}{\lambda}$:$\frac{1}{3y}(1 - \frac{x}{5}) = \frac{1}{5x}(\frac{y}{3} + 1)$$5x(1 - \frac{x}{5}) = 3y(\frac{y}{3} + 1)$$5x - x^{2} = y^{2} + 3y$$x^{2} + y^{2} - 5x + 3y = 0$.

$A$ line cuts the $X$-axis at $A(5,0)$ and the $Y$-axis at $B(0,-3)$. $A$ variable line $PQ$ is drawn perpendicular to $AB$ cutting the $X$-axis at $P$ and the $Y$-axis at $Q$. If $AQ$ and $BP$ meet at $R$, then the locus of $R$ is

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