Question

The coordinates of any point on the circle through the points $A\left( 2,2 \right)$, $B\left( 5,3 \right)$ and $C\left( 3,-1 \right)$ can be written in the form $\left( 4+\sqrt{5}\cos \theta ,1+\sqrt{5}\sin \theta \right)$. Then the coordinates of the point$P$ on $BC$ such that $AP$ is perpendicular to $BC$ are
(a)$\left( -1,4 \right)$
(b)$\left( 4,1 \right)$
(c) $\left( 1,4 \right)$
(d)$\left( 2,3 \right)$

Answer 1

Hint: To find the coordinates of point $P$ find the equation of line joining any two points and assume any point $P$ on the line. Use the fact that the product of slopes of two perpendicular lines is $-1$.

We have the points $A\left( 2,2 \right)$, $B\left( 5,3 \right)$ and $C(3,-1)$ on the circle. Other points on the circle are of the form $\left( 4+\sqrt{5}\cos \theta ,1+\sqrt{5}\sin \theta \right)$. We want to find the coordinates of point $P$ such that the line $AP$ is perpendicular to the line $BC$.
We will begin by finding the equation of line $BC$.
We know that the equation of line joining any two points $(a,b)$ and $\left( c,d \right)$ is $\left( y-b \right)=\dfrac{d-b}{c-a}\left( x-a \right)$.
To find the equation of line $BC$, we will substitute $a=5,b=3,c=3,d=-1$ in the above formula.
Thus, we have $y-3=\dfrac{-1-3}{3-5}\left( x-5 \right)$ as the equation of line $BC$.

& \Rightarrow y-3=2\left( x-5 \right) \\\ & \Rightarrow y-3=2x-10 \\\ & \Rightarrow y=2x-7 \\\ \end{aligned}$$ Hence, the equation of line $$BC$$ is $$y=2x-7$$. $$...\left( 1 \right)$$ Now, we will find the point $$P$$. Let’s assume $$P$$ has coordinates $$(u,v)$$. Substituting $$(u,v)$$ in the equation of line, we have $$v=2u-7$$. $$...\left( 2 \right)$$ To find the slope of line $$AP$$, we will substitute $$a=2,b=2,c=u,d=2u-7$$ in the formula $$\left( \dfrac{d-b}{c-a} \right)$$ of the slope of line. Thus, we have $$\dfrac{2u-7-2}{u-2}=\dfrac{2u-9}{u-2}$$ as the slope of $$AP$$. We know that the product of slopes of two perpendicular lines is $$-1$$. Thus, as $$AP$$ and $$BC$$ are perpendicular to each other, the product of their slopes is $$-1$$. Using equation $$\left( 1 \right)$$ and $$\left( 2 \right)$$, we have $$2\left( \dfrac{2u-9}{u-2} \right)=-1$$ as slope of line $$BC$$ is $$-1$$. On further solving the equation, we have $$4u-18=-u+2$$. $$\begin{aligned} & \Rightarrow 5u=20 \\\ & \Rightarrow u=4 \\\ \end{aligned}$$ Substituting the above value in equation $$\left( 2 \right)$$, we get $$v=2u-7=2\left( 4 \right)-7=8-7=1$$. Thus, the coordinates of $$P$$ are $$(u,v)=\left( 4,1 \right)$$. Hence, the correct answer is $$\left( 4,1 \right)$$. So, the answer is Option (b) Note: One must observe that the point $$P$$ is on the line $$BC$$ and not on the circle. If you take point $$P$$ on the circle, you will get an incorrect answer.