Search any question & find its solution
Question:
Answered & Verified by Expert
Let $A(1,3)$ and $B(2,5)$ be two points and $C(h, k)$ be a point such that $\mathrm{BC}$ is perpendicular to $\mathrm{AC}$. If $\angle \mathrm{CAB}=\angle \mathrm{CBA}$, then $\mathrm{h}=$
Options:
Solution:
1818 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{2}$ or $\frac{5}{2}$
$$
\begin{aligned}
& \text { } A C^2=B C^2 \\
& (h-1)^2+(k-3)^2=(h-2)^2+(k-5)^2 \\
& \Rightarrow h^2+k^2-2 h-6 k+10=h^2+k^2-4 h-10 k+29 \\
& \Rightarrow 2 h+4 k=19
\end{aligned}
$$
Also $A C \perp B C$
$\begin{aligned} & m_{A C} \times m_{B C}=-1 \\ & \Rightarrow \frac{k-3}{h-1} \times \frac{k-5}{h-2}=-1 \\ & \Rightarrow(k-3)(k-5)=-(h-1)(h-2) \\ & \Rightarrow k^2-8 k+15=-h^2+3 h-2 \\ & \Rightarrow h^2+k^2-3 h-8 k+17=0 \\ & \because \text { from }(1), k=\frac{19-2 h}{4} \\ & \Rightarrow h^2+\left(\frac{19-2 h}{4}\right)^2-3 h-8\left(\frac{19-2 h}{4}\right)+17=0 \\ & \Rightarrow 16 h^2+(19-2 h)^2-48 h-32(19-2 h)+272=0 \\ & \Rightarrow 16 h^2+361+4 h^2-76 h-48 h-608+64 h+272=0 \\ & \Rightarrow 20 h^2-60 h+25=0 \Rightarrow 4 h^2-12 h+5=0 \\ & \Rightarrow 4 h^2-10 h-2 h+5=0 \Rightarrow(2 h-1)(2 h-5)=0 \\ & \therefore h=\frac{1}{2}, \frac{5}{2} .\end{aligned}$
\begin{aligned}
& \text { } A C^2=B C^2 \\
& (h-1)^2+(k-3)^2=(h-2)^2+(k-5)^2 \\
& \Rightarrow h^2+k^2-2 h-6 k+10=h^2+k^2-4 h-10 k+29 \\
& \Rightarrow 2 h+4 k=19
\end{aligned}
$$
Also $A C \perp B C$
$\begin{aligned} & m_{A C} \times m_{B C}=-1 \\ & \Rightarrow \frac{k-3}{h-1} \times \frac{k-5}{h-2}=-1 \\ & \Rightarrow(k-3)(k-5)=-(h-1)(h-2) \\ & \Rightarrow k^2-8 k+15=-h^2+3 h-2 \\ & \Rightarrow h^2+k^2-3 h-8 k+17=0 \\ & \because \text { from }(1), k=\frac{19-2 h}{4} \\ & \Rightarrow h^2+\left(\frac{19-2 h}{4}\right)^2-3 h-8\left(\frac{19-2 h}{4}\right)+17=0 \\ & \Rightarrow 16 h^2+(19-2 h)^2-48 h-32(19-2 h)+272=0 \\ & \Rightarrow 16 h^2+361+4 h^2-76 h-48 h-608+64 h+272=0 \\ & \Rightarrow 20 h^2-60 h+25=0 \Rightarrow 4 h^2-12 h+5=0 \\ & \Rightarrow 4 h^2-10 h-2 h+5=0 \Rightarrow(2 h-1)(2 h-5)=0 \\ & \therefore h=\frac{1}{2}, \frac{5}{2} .\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.