MHT CET · Maths · Three Dimensional Geometry
Let \(a, b \in R\). If the mirror image of the point \(\mathrm{p}(\mathrm{a}, 6,9)\) w.r.t. 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 90
- D 84
Answer & Solution
Correct Answer
(A) 88
Step-by-step Solution
Detailed explanation
Note that mid-point of the line joining points \((a, 6,9)\) and \((20, b,-a-9)\) lies on the given line.
\(\therefore \quad\) The midpoint is \(\left(\frac{\mathrm{a}+20}{2}, \frac{6+\mathrm{b}}{2}, \frac{9-\mathrm{a}-9}{2}\right)\)
Substituting this point in the equation of the given line, we get
\(\begin{array}{ll}
& \frac{\frac{a+20}{2}-3}{7}=\frac{\frac{6+b}{2}-2}{5}=\frac{\frac{-a}{2}-1}{-9} \\
\therefore \quad & \frac{a+14}{14}=\frac{a+2}{18} \\
\therefore \quad & a=-56 \\
\therefore \quad & \frac{-56+4}{14}=\frac{2+b}{10} \\
\therefore \quad & b=-32 \\
\therefore \quad & |a+b|=88
\end{array}\)
\(\therefore \quad\) The midpoint is \(\left(\frac{\mathrm{a}+20}{2}, \frac{6+\mathrm{b}}{2}, \frac{9-\mathrm{a}-9}{2}\right)\)
Substituting this point in the equation of the given line, we get
\(\begin{array}{ll}
& \frac{\frac{a+20}{2}-3}{7}=\frac{\frac{6+b}{2}-2}{5}=\frac{\frac{-a}{2}-1}{-9} \\
\therefore \quad & \frac{a+14}{14}=\frac{a+2}{18} \\
\therefore \quad & a=-56 \\
\therefore \quad & \frac{-56+4}{14}=\frac{2+b}{10} \\
\therefore \quad & b=-32 \\
\therefore \quad & |a+b|=88
\end{array}\)
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