If P(6,1) be the orthocentre of the triangle whose vertices are A(5,-2),B(8,3) and C(h,k),…

Question

If $P(6,1)$ be the orthocentre of the triangle whose vertices are $A(5,-2),B(8,3)$ and $C(h,k)$, then the point $C$ lies on the circle :

Accepted Answer

Slope of $$B F=\left(m_{B F}\right)=\frac{3-1}{8-6}=1$$

$$\begin{aligned}
& m_{A C}=\frac{k+2}{h-5}=-1(A C \perp B F) \
& \Rightarrow k+2=5-h \
& \Rightarrow k=3-h \quad \ldots \text { (i) } \
& m_{A D}=\frac{1+2}{6-5}=3 \
& m_{B C}=\frac{k-3}{h-8}=-\frac{1}{3}(A D \perp B C) \
& \Rightarrow 3 k-9=8-h \quad \ldots \text { (ii) }
\end{aligned}$$

From (i) \& (ii)

$h=-4, k=7$

Now, $h^2+k^2=16+49=65$

$h^2+k^2-65=0$

Locus is $x^2+y^2-65=0$

Solutions · 1