JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let \(a, b \in R\). If the mirror image of the point \(P( a ,6,9)\) with respect to the line \(\frac{x-3}{7}=\frac{y-2}{5}=\frac{z-1}{-9}\) is \((20, b,-a-9),\) then \(|a+b|\) is equal to :
- A \(88\)
- B \(86\)
- C \(84\)
- D \(90\)
Answer & Solution
Correct Answer
(A) \(88\)
Step-by-step Solution
Detailed explanation
\(P (9,6,9)\) \(\frac{ x -3}{7}=\frac{ y -2}{5}=\frac{ z -1}{-9}\) \(Q =(20, b ,- a -9)\) \(\frac{\frac{20+a}{2}-3}{7}=\frac{\frac{b+6}{2}-2}{5}=\frac{-\frac{9}{2}-1}{-9}\) \(\frac{14+9}{14}=\frac{b+2}{10}=\frac{a+2}{18}\) \(\Rightarrow a=-56\) and \(b=-32\)…
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