Let A(1, 3) and B(2, 5) be two points and C(h | vedclass

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(C) Given that $\angle CAB = \angle CBA$,the triangle $	riangle ABC$ is an isosceles triangle with $AC = BC$.Squaring both sides,we get $AC^2 = BC^2$

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$\frac{1}{2}$ or $\frac{5}{2}$

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(C) Given that $\angle CAB = \angle CBA$,the triangle $	riangle ABC$ is an isosceles triangle with $AC = BC$.Squaring both sides,we get $AC^2 = BC^2$.$(h-1)^2 + (k-3)^2 = (h-2)^2 + (k-5)^2$$h^2 - 2h + 1 + k^2 - 6k + 9 = h^2 - 4h + 4 + k^2 - 10k + 25$$-2h - 6k + 10 = -4h - 10k + 29$$2h + 4k = 19 \implies k = \frac{19-2h}{4} \quad (1)$Since $AC \perp BC$,the product of their slopes is $-1$:$m_{AC} 	imes m_{BC} = -1$$\frac{k-3}{h-1} 	imes \frac{k-5}{h-2} = -1$$(k-3)(k-5) = -(h-1)(h-2)$$k^2 - 8k + 15 = -(h^2 - 3h + 2) = -h^2 + 3h - 2$$h^2 + k^2 - 3h - 8k + 17 = 0 \quad (2)$Substituting $(1)$ into $(2)$:$h^2 + (\frac{19-2h}{4})^2 - 3h - 8(\frac{19-2h}{4}) + 17 = 0$$16h^2 + (361 - 76h + 4h^2) - 48h - 32(19-2h) + 272 = 0$$20h^2 - 124h + 361 - 608 + 64h + 272 = 0$$20h^2 - 60h + 25 = 0$$4h^2 - 12h + 5 = 0$$(2h-1)(2h-5) = 0$$h = \frac{1}{2}$ or $h = \frac{5}{2}$.

Let $A(1, 3)$ and $B(2, 5)$ be two points and $C(h, k)$ be a point such that $BC$ is perpendicular to $AC$. If $\angle CAB = \angle CBA$,then $h =$

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