A line cuts the x-axis at A(7, 0) and the y-a | vedclass

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(C) The slope of line $AB$ is $m_{AB} = \frac{-5 - 0}{0 - 7} = \frac{5}{7}$.Since $PQ \perp AB$, the slope of line $PQ$ is $m_{PQ} = -\frac{7}{5}$.The

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$x^{2}+y^{2}-7x+5y=0$

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(C) The slope of line $AB$ is $m_{AB} = \frac{-5 - 0}{0 - 7} = \frac{5}{7}$.Since $PQ \perp AB$, the slope of line $PQ$ is $m_{PQ} = -\frac{7}{5}$.The equation of line $PQ$ is $y - 0 = -\frac{7}{5}(x - a)$, which simplifies to $7x + 5y = 7a$.Since $Q(0, b)$ lies on $PQ$, $5b = 7a$, so $b = \frac{7a}{5}$.$R(h, k)$ is the intersection of $AQ$ and $BP$.The equation of line $AQ$ passing through $A(7, 0)$ and $Q(0, b)$ is $\frac{x}{7} + \frac{y}{b} = 1$.The equation of line $BP$ passing through $B(0, -5)$ and $P(a, 0)$ is $\frac{x}{a} + \frac{y}{-5} = 1$.Since $R(h, k)$ lies on both lines:$(1)$ $\frac{h}{7} + \frac{k}{b} = 1$ $\Rightarrow \frac{h}{7} + \frac{5k}{7a} = 1$ $\Rightarrow ah + 5k = 7a$ $\Rightarrow a(7 - h) = 5k$ $\Rightarrow a = \frac{5k}{7 - h}$.$(2)$ $\frac{h}{a} - \frac{k}{5} = 1$ $\Rightarrow 5h - ak = 5a$ $\Rightarrow 5h = a(k + 5)$ $\Rightarrow a = \frac{5h}{k + 5}$.Equating the two expressions for $a$:$\frac{5k}{7 - h} = \frac{5h}{k + 5}$ $\Rightarrow k(k + 5) = h(7 - h)$ $\Rightarrow k^2 + 5k = 7h - h^2$.Rearranging gives $h^2 + k^2 - 7h + 5k = 0$.Replacing $(h, k)$ with $(x, y)$, the locus is $x^2 + y^2 - 7x + 5y = 0$.

$A$ 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 at $P(a, 0)$ and the $y$-axis at $Q(0, b)$. If $AQ$ and $BP$ intersect at $R(h, k)$, the locus of $R$ is

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