Two vertices of a triangle ABC are A(3,-1) and B(-2,3) , and its orthocentre is P(1,1) . If the coordinates of the point C are (α, β) and the centre of the of the circle circumscribing the triangle PAB is (h, k) , then

Correct answer: A. 5 M_ AB =(4)/(-5) M_ DP =(5)/(4) Equation of PC is y-1=(5)/(4)(x-1) (1) M_ AP =(2)/(-2)=-1 M_ BC =+1 Equation of BC is y-3=(x+2) (2) On solving (1) and (2) aligned & x+4=(5)/(4)(x-1) 4 x+16=5 x-5 α=21 & β=y=x+5=26 & α+β=47 aligned Equation of bisector of AP y-0=(x-2) (3) Equation of bisector of AB y-1=(5)/(4)(x-(1)/(2)) (4) aligned & On solving (3) & (4) & (x-3) 4=5 x-(5)/(2) & x=(-19)/(2)=h & y=(-23)/(2)=k & 2( h+k)=-42 aligned

JEE Main 20249 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 9 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Straight Lines

JEE Main 2024 · 9 April · Evening Shift · Mathematics · Straight Lines

Two vertices of a triangle ABC are A(3,-1) and B(-2,3) , and its...

Two vertices of a triangle

\mathrm{ABC}

are

\mathrm{A}(3,-1)

and

\mathrm{B}(-2,3)

, and its orthocentre is

\mathrm{P}(1,1)

. If the coordinates of the point

\mathrm{C}

are

(\alpha, \beta)

and the centre of the of the circle circumscribing the triangle

\mathrm{PAB}

(\mathrm{h}, \mathrm{k})

, then the value of

(\alpha+\beta)+2(\mathrm{~h}+\mathrm{k})

equals

Options

Correct answer

A. 5

Step-by-step solution

\mathrm{M}_{\mathrm{AB}}=\frac{4}{-5} \Rightarrow \mathrm{M}_{\mathrm{DP}}=\frac{5}{4}

Equation of

\mathrm{PC}

\mathrm{y}-1=\frac{5}{4}(\mathrm{x}-1) \ldots(1) \mathrm{M}_{\mathrm{AP}}=\frac{2}{-2}=-1 \Rightarrow \mathrm{M}_{\mathrm{BC}}=+1

Equation of

\mathrm{BC}

\mathrm{y}-3=(\mathrm{x}+2)\ldots(2)

On solving (1) and (2)

\begin{aligned} & x+4=\frac{5}{4}(x-1) \Rightarrow 4 x+16=5 x-5 \Rightarrow \alpha=21 \\ & \Rightarrow \beta=y=x+5=26 \\ & \alpha+\beta=47 \end{aligned}

Equation of

\perp

bisector of AP

y-0=(x-2)\ldots(3)

Equation of

\perp

bisector of

\mathrm{AB} y-1=\frac{5}{4}\left(x-\frac{1}{2}\right)\ldots(4)

\begin{aligned} & \text{ On solving (3) & (4) } \\ & (\mathrm{x}-3) 4=5 \mathrm{x}-\frac{5}{2} \\ & \mathrm{x}=\frac{-19}{2}=\mathrm{h} \\ & \mathrm{y}=\frac{-23}{2}=\mathrm{k} \\ & \Rightarrow 2(\mathrm{ h}+\mathrm{k})=-42\end{aligned}

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