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 circle circumscribing the triangle PAB is (h, k) , then the value of (α+β)+2(h+k) equals :

Mathematics · Straight lines

JEE Main 2024 — 9 April, Shift 2 — Question 7

Two vertices of a triangle 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 $C$ are $(\alpha, \beta)$ and the centre of the circle circumscribing the triangle PAB is $(h, k)$ , then the value of $(\alpha+\beta)+2(h+k)$ equals :

Option A:
51
Option B:
81
Option C:
5
Correct
Option D:
15

Answer: C

Step-by-step solution

\mathrm{Given:}\quad A(3,-1),\; B(-2,3),\; P(1,1)\ (\mathrm{orthocentre\ of\ } \triangle ABC).

\mathrm{Let}\ C=(\alpha,\beta).

\mathrm{Since\ }P\mathrm{\ lies\ on\ the\ altitude\ from\ }A,\ \mathrm{slope}(AP)\cdot \mathrm{slope}(BC)=-1.

\mathrm{slope}(AP)=\frac{1-(-1)}{1-3}=\frac{2}{-2}=-1.

\therefore\ \mathrm{slope}(BC)=1\quad\Rightarrow\quad \frac{\beta-3}{\alpha+2}=1.

\therefore\ \beta-3=\alpha+2\quad\Rightarrow\quad \beta=\alpha+5.

\mathrm{Since\ }P\mathrm{\ lies\ on\ the\ altitude\ from\ }B,\ \mathrm{slope}(BP)\cdot \mathrm{slope}(AC)=-1.

\mathrm{slope}(BP)=\frac{1-3}{1-(-2)}=\frac{-2}{3}=-\tfrac{2}{3}.

\therefore\ \mathrm{slope}(AC)=\frac{3}{2}\quad\Rightarrow\quad \frac{\beta-(-1)}{\alpha-3}=\frac{3}{2}.

\mathrm{Hence}\quad \beta+1=\tfrac{3}{2}(\alpha-3)\quad\Rightarrow\quad 2\beta+2=3\alpha-9.

\mathrm{Substitute}\ \beta=\alpha+5:\quad 2(\alpha+5)+2=3\alpha-9.

2\alpha+10+2=3\alpha-9\quad\Rightarrow\quad 12= \alpha-9\quad\Rightarrow\quad \alpha=21.

\mathrm{Thus}\quad \beta=\alpha+5=26\quad\Rightarrow\quad C=(21,26).

\mathrm{Now\ find\ circumcentre\ of\ } \triangle PAB.

\mathrm{Midpoint\ of\ }PA:\ M_1=\bigl(\tfrac{1+3}{2},\tfrac{1+(-1)}{2}\bigr)=(2,0).

\mathrm{slope}(PA)=-1\quad\Rightarrow\quad \mathrm{slope\ of\ perpendicular\ bisector}=1.

\mathrm{Equation\ of\ perpendicular\ bisector\ through\ }M_1:\quad y=x-2.

\mathrm{Midpoint\ of\ }PB:\ M_2=\bigl(\tfrac{1+(-2)}{2},\tfrac{1+3}{2}\bigr)=\bigl(-\tfrac{1}{2},2\bigr).

\mathrm{slope}(PB)=-\tfrac{2}{3}\quad\Rightarrow\quad \mathrm{slope\ of\ perpendicular\ bisector}=\tfrac{3}{2}.

\mathrm{Equation\ through\ }M_2:\quad y-2=\tfrac{3}{2}\bigl(x+\tfrac{1}{2}\bigr) \quad\Rightarrow\quad y=\tfrac{3}{2}x+\tfrac{11}{4}.

\mathrm{Intersect\ }y=x-2\ \mathrm{and}\ y=\tfrac{3}{2}x+\tfrac{11}{4}: \quad x-2=\tfrac{3}{2}x+\tfrac{11}{4}.

-x\cdot\tfrac{1}{2}=\tfrac{11}{4}+2=\tfrac{19}{4}\quad\Rightarrow\quad x=-\tfrac{19}{2}.

\mathrm{Then}\quad y=x-2=-\tfrac{19}{2}-2=-\tfrac{23}{2}.

\mathrm{Thus\ the\ circumcentre\ }(h,k)=\Bigl(-\tfrac{19}{2},-\tfrac{23}{2}\Bigr).

\mathrm{Compute}\quad (\alpha+\beta)+2(h+k).

\alpha+\beta=21+26=47,\qquad h+k=-\tfrac{19}{2}-\tfrac{23}{2}=-21.

2(h+k)=-42.

\therefore\quad (\alpha+\beta)+2(h+k)=47-42=5.

\boxed{5}

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 9 April, Shift 2
Subject: Mathematics
Chapter: Straight lines
Topic: Various forms of lines