MCQMediumJEE 2023Conic Sections (Parabola, Ellipse, Hyperbola)

JEE Mathematics 2023 Question with Solution

If (α,β)(\alpha,\beta) is the orthocenter of the triangle ABCABC with vertices A(3,7),B(1,2),C(4,5)A(3,-7),\quad B(-1,2),\quad C(4,5), then 9α6β+609\alpha-6\beta+60 is equal to

  • A

    2525

  • B

    3030

  • C

    3535

  • D

    4040

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The triangle has vertices A(3,7)A(3,-7), B(1,2)B(-1,2) and C(4,5)C(4,5).

Find: The value of 9α6β+609\alpha-6\beta+60, where (α,β)(\alpha,\beta) is the orthocenter.

Find the slopes of two sides and then use perpendicular slopes to write the equations of two altitudes.

Slope of BCBC is

mBC=524(1)=35m_{BC}=\frac{5-2}{4-(-1)}=\frac{3}{5}

So, the slope of the altitude from AA is

m1=53m_1=-\frac{5}{3}

Slope of ACAC is

mAC=5(7)43=12m_{AC}=\frac{5-(-7)}{4-3}=12

So, the slope of the altitude from BB is

m2=112m_2=-\frac{1}{12}

Equation of altitude from A(3,7)A(3,-7):

y+7=53(x3)y+7=-\frac{5}{3}(x-3) 3y+21=5x+153y+21=-5x+15 5x+3y+6=05x+3y+6=0

Equation of altitude from B(1,2)B(-1,2):

y2=112(x+1)y-2=-\frac{1}{12}(x+1) 12y24=x112y-24=-x-1 x+12y23=0x+12y-23=0

The orthocenter is the intersection of these two altitudes. So solve

5x+3y=65x+3y=-6 x+12y=23x+12y=23

Multiply the second equation by 55:

5x+60y=1155x+60y=115

Subtract the first equation:

57y=12157y=121 y=12157y=\frac{121}{57}

Substitute back into x+12y=23x+12y=23:

x+1212157=23x+12\cdot\frac{121}{57}=23 x=5723145257=4757x=\frac{57\cdot 23-1452}{57}=-\frac{47}{57}

Thus,

α=4757,β=12157\alpha=-\frac{47}{57},\quad \beta=\frac{121}{57}

Now compute the required expression:

9α6β+60=9(4757)6(12157)+609\alpha-6\beta+60=9\left(-\frac{47}{57}\right)-6\left(\frac{121}{57}\right)+60 =423726+342057=227157=25=\frac{-423-726+3420}{57}=\frac{2271}{57}=25

Therefore, the value of the expression is 2525, so the correct option is A.

Use Two Altitudes Only

Given: The orthocenter is the intersection point of altitudes.

Find: The required linear expression in α\alpha and β\beta.

A shortcut here is that only two altitudes are needed to locate the orthocenter. There is no need to use all three sides.

  • Use side BCBC to get the altitude from AA.
  • Use side ACAC to get the altitude from BB.
  • Solve the two linear equations directly.

This works because all altitudes of a triangle are concurrent at the orthocenter.

From the working:

5x+3y+6=05x+3y+6=0 x+12y23=0x+12y-23=0

Their intersection gives

(α,β)=(4757,12157)(\alpha,\beta)=\left(-\frac{47}{57},\frac{121}{57}\right)

Then substitute into

9α6β+609\alpha-6\beta+60

to get 2525. Hence the correct option is A.

Common mistakes

  • Using the same slope for the altitude as for the side. This is wrong because an altitude is perpendicular to the side, so its slope must be the negative reciprocal. Use m=1mm_\perp=-\frac{1}{m} when the side slope is nonzero.

  • Making a sign error while computing slopes such as 5(7)43\frac{5-(-7)}{4-3}. This changes the altitude equation and gives a wrong orthocenter. Carefully simplify numerator and denominator before taking the reciprocal.

  • Solving the two altitude equations incorrectly after forming them. The orthocenter is the intersection point, so both linear equations must be satisfied simultaneously. Use elimination or substitution carefully before evaluating 9α6β+609\alpha-6\beta+60.

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