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 (\alpha, \beta) and the centre of the of the circle circumscribing the triangle PAB is (h, k) , then the value of (\alpha+\beta)+2(~h+k) equals

Properties of Triangle PYQ — JEE Main 2024 · 9 Apr · Shift 2 · Q35 with Solution

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Correct answer: D

Given data

We have:

$A(3,-1)$
$B(-2,3)$
Orthocentre of $\triangle ABC$ is $P(1,1)$
$C=(\alpha,\beta)$

We need:

(\alpha+\beta)+2(h+k)

where $(h,k)$ is the circumcentre of triangle $PAB$ .

Use orthocentre property to find point $C$

Since $P$ is the orthocentre of $\triangle ABC$ :

$AP \perp BC$
$BP \perp AC$

Step 2.1: Slope of $AP$

Points $A(3,-1)$ and $P(1,1)$ give

m_{AP}=\frac{1-(-1)}{1-3}=\frac{2}{-2}=-1

So line $BC$ has slope

m_{BC}=1

Since $B(-2,3)$ lies on $BC$ ,

y-3=1(x+2)

y=x+5

So $C=(\alpha,\beta)$ lies on

\beta=\alpha+5

Step 2.2: Slope of $BP$

Points $B(-2,3)$ and $P(1,1)$ give

m_{BP}=\frac{1-3}{1-(-2)}=\frac{-2}{3}

So line $AC$ has slope

m_{AC}=\frac{3}{2}

Since $A(3,-1)$ lies on $AC$ ,

y+1=\frac{3}{2}(x-3)

y=\frac{3}{2}x-\frac{11}{2}

So $C$ also lies on this line.

Step 2.3: Intersection gives $C$

Solve

y=x+5

and

y=\frac{3}{2}x-\frac{11}{2}

Equating,

x+5=\frac{3}{2}x-\frac{11}{2}

Multiply by $2$ :

2x+10=3x-11

x=21

Then

y=x+5=26

Hence,

C=(21,26)

So,

\alpha=21,\quad \beta=26

Thus,

\alpha+\beta=47

Find circumcentre of triangle $PAB$

We need the circumcentre of triangle with vertices:

$P(1,1)$
$A(3,-1)$
$B(-2,3)$

The circumcentre is the intersection of perpendicular bisectors.

Step 3.1: Perpendicular bisector of $AP$

Midpoint of $AP$ :

M_1=\left(\frac{3+1}{2},\frac{-1+1}{2}\right)=(2,0)

Slope of $AP$ is $-1$ , so perpendicular slope is $1$ .

Equation:

y-0=1(x-2)

y=x-2

Step 3.2: Perpendicular bisector of $AB$

Midpoint of $AB$ :

M_2=\left(\frac{3+(-2)}{2},\frac{-1+3}{2}\right)=\left(\frac12,1\right)

Slope of $AB$ :

m_{AB}=\frac{3-(-1)}{-2-3}=\frac{4}{-5}=-\frac45

So perpendicular slope is

\frac54

Equation of perpendicular bisector:

y-1=\frac54\left(x-\frac12\right)

Step 3.3: Solve with $y=x-2$

Substitute $y=x-2$ :

x-2-1=\frac54\left(x-\frac12\right)

x-3=\frac54x-\frac58

Multiply by $8$ :

8x-24=10x-5

-19=2x

x=-\frac{19}{2}

Then

y=x-2=-\frac{19}{2}-2=-\frac{23}{2}

Therefore,

(h,k)=\left(-\frac{19}{2},-\frac{23}{2}\right)

So,

h+k=-\frac{19}{2}-\frac{23}{2}=-21

Hence,

2(h+k)=-42

Compute required value

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

Match with options

The value is

\boxed{5}

which is Option D.

Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So the stored answer agrees.

More from Properties of Triangle

View written solutionFree

Correct answer: D

Given data

We have:

$A(3,-1)$
$B(-2,3)$
Orthocentre of $\triangle ABC$ is $P(1,1)$
$C=(\alpha,\beta)$

We need:

(\alpha+\beta)+2(h+k)

where $(h,k)$ is the circumcentre of triangle $PAB$ .

Use orthocentre property to find point $C$

Since $P$ is the orthocentre of $\triangle ABC$ :

$AP \perp BC$
$BP \perp AC$

Step 2.1: Slope of $AP$

Points $A(3,-1)$ and $P(1,1)$ give

m_{AP}=\frac{1-(-1)}{1-3}=\frac{2}{-2}=-1

So line $BC$ has slope

m_{BC}=1

Since $B(-2,3)$ lies on $BC$ ,

y-3=1(x+2)

y=x+5

So $C=(\alpha,\beta)$ lies on

\beta=\alpha+5

Step 2.2: Slope of $BP$

Points $B(-2,3)$ and $P(1,1)$ give

m_{BP}=\frac{1-3}{1-(-2)}=\frac{-2}{3}

So line $AC$ has slope

m_{AC}=\frac{3}{2}

Since $A(3,-1)$ lies on $AC$ ,

y+1=\frac{3}{2}(x-3)

y=\frac{3}{2}x-\frac{11}{2}

So $C$ also lies on this line.

Step 2.3: Intersection gives $C$

Solve

y=x+5

and

y=\frac{3}{2}x-\frac{11}{2}

Equating,

x+5=\frac{3}{2}x-\frac{11}{2}

Multiply by $2$ :

2x+10=3x-11

x=21

Then

y=x+5=26

Hence,

C=(21,26)

So,

\alpha=21,\quad \beta=26

Thus,

\alpha+\beta=47

Find circumcentre of triangle $PAB$

We need the circumcentre of triangle with vertices:

$P(1,1)$
$A(3,-1)$
$B(-2,3)$

The circumcentre is the intersection of perpendicular bisectors.

Step 3.1: Perpendicular bisector of $AP$

Midpoint of $AP$ :

M_1=\left(\frac{3+1}{2},\frac{-1+1}{2}\right)=(2,0)

Slope of $AP$ is $-1$ , so perpendicular slope is $1$ .

Equation:

y-0=1(x-2)

y=x-2

Step 3.2: Perpendicular bisector of $AB$

Midpoint of $AB$ :

M_2=\left(\frac{3+(-2)}{2},\frac{-1+3}{2}\right)=\left(\frac12,1\right)

Slope of $AB$ :

m_{AB}=\frac{3-(-1)}{-2-3}=\frac{4}{-5}=-\frac45

So perpendicular slope is

\frac54

Equation of perpendicular bisector:

y-1=\frac54\left(x-\frac12\right)

Step 3.3: Solve with $y=x-2$

Substitute $y=x-2$ :

x-2-1=\frac54\left(x-\frac12\right)

x-3=\frac54x-\frac58

Multiply by $8$ :

8x-24=10x-5

-19=2x

x=-\frac{19}{2}

Then

y=x-2=-\frac{19}{2}-2=-\frac{23}{2}

Therefore,

(h,k)=\left(-\frac{19}{2},-\frac{23}{2}\right)

So,

h+k=-\frac{19}{2}-\frac{23}{2}=-21

Hence,

2(h+k)=-42

Compute required value

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

Match with options

The value is

\boxed{5}

which is Option D.

Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So the stored answer agrees.

Properties of Triangle question

Properties of Triangle question

Step 2.1: Slope of $AP$

Step 2.2: Slope of $BP$

Step 2.3: Intersection gives $C$

Step 3.1: Perpendicular bisector of $AP$

Step 3.2: Perpendicular bisector of $AB$

Step 3.3: Solve with $y=x-2$

More from Properties of Triangle

Properties of Triangle question

Properties of Triangle question

Step 2.1: Slope of $AP$

Step 2.2: Slope of $BP$

Step 2.3: Intersection gives $C$

Step 3.1: Perpendicular bisector of $AP$

Step 3.2: Perpendicular bisector of $AB$

Step 3.3: Solve with $y=x-2$

More from Properties of Triangle

Step 2.1: Slope of APAPAP

Step 2.2: Slope of BPBPBP

Step 2.3: Intersection gives CCC

Step 3.1: Perpendicular bisector of APAPAP

Step 3.2: Perpendicular bisector of ABABAB

Step 3.3: Solve with y=x−2y=x-2y=x−2

More from Properties of Triangle

Step 2.1: Slope of APAPAP

Step 2.2: Slope of BPBPBP

Step 2.3: Intersection gives CCC

Step 3.1: Perpendicular bisector of APAPAP

Step 3.2: Perpendicular bisector of ABABAB

Step 3.3: Solve with y=x−2y=x-2y=x−2

More from Properties of Triangle

Step 2.1: Slope of $AP$

Step 2.2: Slope of $BP$

Step 2.3: Intersection gives $C$

Step 3.1: Perpendicular bisector of $AP$

Step 3.2: Perpendicular bisector of $AB$

Step 3.3: Solve with $y=x-2$

Step 2.1: Slope of $AP$

Step 2.2: Slope of $BP$

Step 2.3: Intersection gives $C$

Step 3.1: Perpendicular bisector of $AP$

Step 3.2: Perpendicular bisector of $AB$

Step 3.3: Solve with $y=x-2$