# 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)\[\left( 2,3 \right)\]

Last updated date: 29th Mar 2023

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Answer

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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 \[BC\].

\[\begin{align}

& \Rightarrow y-3=2\left( x-5 \right) \\

& \Rightarrow y-3=2x-10 \\

& \Rightarrow y=2x-7 \\

\end{align}\]

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{align}

& \Rightarrow 5u=20 \\

& \Rightarrow u=4 \\

\end{align}\]

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.

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 \[BC\].

\[\begin{align}

& \Rightarrow y-3=2\left( x-5 \right) \\

& \Rightarrow y-3=2x-10 \\

& \Rightarrow y=2x-7 \\

\end{align}\]

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{align}

& \Rightarrow 5u=20 \\

& \Rightarrow u=4 \\

\end{align}\]

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.

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