Verified Solution • 2025–26 Session moderate

QJEEM21Ma1d12: JEE Main 2021 Mathematics - Class 12 JEEM21Ma1d12 Solution: Let a, b R. If the mirror image of the point P(a, 6, 9) with…

Let a, b ∈\in R. If the mirror image of the point P(a, 6, 9) with respect to the line

x−37=y−25=z−1−9{{x - 3} \over 7} = {{y - 2} \over 5} = {{z - 1} \over { - 9}} is (20, b, −- a −- 9), then | a + b |, is equal to :

Multiple Choice Options

  • A 88
  • B 90
  • C 86
  • D 84

Step-by-Step Verified Solution

Given, P(a, 6, 9)

Equation of line x−37=y−25=z−1−9{{x - 3} \over 7} = {{y - 2} \over 5} = {{z - 1} \over { - 9}}

Image of point P with respect to line is point Q(20, b, −- a −- 9)

Mid-point of P and Q = (a+202,6+b2,−a2)\left( {{{a + 20} \over 2},{{6 + b} \over 2},{{ - a} \over 2}} \right)

This point lies on line

∴\therefore a+202−37=6+b2−25=−a2−1−9{{{{a + 20} \over 2} - 3} \over 7} = {{{{6 + b} \over 2} - 2} \over 5} = {{{{ - a} \over 2} - 1} \over { - 9}}

⇒a+1414=b+210=a+218\Rightarrow {{a + 14} \over {14}} = {{b + 2} \over {10}} = {{a + 2} \over {18}}

⇒a+1414=a+218\Rightarrow {{a + 14} \over {14}} = {{a + 2} \over {18}} and b+210=a+218{{b + 2} \over {10}} = {{a + 2} \over {18}}

Solving, we get a = −- 56, b = −- 32

∴\therefore ∣a+b∣=∣−56−32∣=88\left| {a + b} \right| = \left| { - 56 - 32} \right| = 88

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