Search any question & find its solution
Question:
Answered & Verified by Expert
The reflection of the point \((4,-13)\) in the line \(5 x+y+6=0\), is
Options:
Solution:
1008 Upvotes
Verified Answer
The correct answer is:
\((-1,-14)\)
Let \(\mathrm{Q}(a, b)\) be the reflection of \(\mathrm{P}(4,-13)\) in the line \(5 x+y+6=0\)
Then the mid-point \(R\left(\frac{a+4}{2}, \frac{b-13}{2}\right)\) lies on
\(\begin{aligned}
& 5 x+y+6=0 \\
& \therefore 5\left(\frac{a+4}{2}\right)+\frac{b-13}{2}+6=0 \\
& \Rightarrow 5 a+b+19=0 \quad ...(i)
\end{aligned}\)
Also \(\mathrm{PQ}\) is perpendicular to \(5 x+y+6=0\)
Therefore \(\frac{b+13}{a-4} \times\left(-\frac{5}{1}\right)=-1\)
\(\Rightarrow a-5 b-69=0 \quad ...(ii)\)
Solving (i) and (ii), we get \(a=-1, b=-14\)
Then the mid-point \(R\left(\frac{a+4}{2}, \frac{b-13}{2}\right)\) lies on
\(\begin{aligned}
& 5 x+y+6=0 \\
& \therefore 5\left(\frac{a+4}{2}\right)+\frac{b-13}{2}+6=0 \\
& \Rightarrow 5 a+b+19=0 \quad ...(i)
\end{aligned}\)
Also \(\mathrm{PQ}\) is perpendicular to \(5 x+y+6=0\)
Therefore \(\frac{b+13}{a-4} \times\left(-\frac{5}{1}\right)=-1\)
\(\Rightarrow a-5 b-69=0 \quad ...(ii)\)
Solving (i) and (ii), we get \(a=-1, b=-14\)
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.