KCET · Maths · Straight Lines
The reflection of the point \((1,1)\) along the line \(y=-x\) is
- A \((0,0)\)
- B \((-1,1)\)
- C \((-1,-1)\)
- D \((1,-1)\)
Answer & Solution
Correct Answer
(C) \((-1,-1)\)
Step-by-step Solution
Detailed explanation
The mid-point of \(P Q\) lies on \(x+y=0\)
\[
\begin{gathered}
\left(\frac{\mathrm{h}+1}{2}\right)+\left(\frac{\mathrm{k}+1}{2}\right)=0 \\
\mathrm{~h}+\mathrm{k}=-2
\end{gathered}
\]
Again, \(\mathrm{PQ}\) is perpendicular to \(\mathrm{AB}\).
\[
\begin{aligned}
& \therefore &\left(\frac{\mathrm{k}-1}{\mathrm{~h}-1}\right)(-1) &=-1 \\
\Rightarrow & &\left(\because \mathrm{m}_{1} \mathrm{~m}_{2}=-1\right) \\
& \Rightarrow & \mathrm{k}-1 &=\mathrm{h}-1 \\
\mathrm{k} &=\mathrm{h}
\end{aligned}
\]
Put \(\mathrm{k}=\mathrm{h}\) in Eq. (i), we get
\[
2 \mathrm{~h}=-2 \Rightarrow \mathrm{h}=1 \quad \therefore \mathrm{k}=-1
\]
\[
\begin{gathered}
\left(\frac{\mathrm{h}+1}{2}\right)+\left(\frac{\mathrm{k}+1}{2}\right)=0 \\
\mathrm{~h}+\mathrm{k}=-2
\end{gathered}
\]
Again, \(\mathrm{PQ}\) is perpendicular to \(\mathrm{AB}\).
\[
\begin{aligned}
& \therefore &\left(\frac{\mathrm{k}-1}{\mathrm{~h}-1}\right)(-1) &=-1 \\
\Rightarrow & &\left(\because \mathrm{m}_{1} \mathrm{~m}_{2}=-1\right) \\
& \Rightarrow & \mathrm{k}-1 &=\mathrm{h}-1 \\
\mathrm{k} &=\mathrm{h}
\end{aligned}
\]
Put \(\mathrm{k}=\mathrm{h}\) in Eq. (i), we get
\[
2 \mathrm{~h}=-2 \Rightarrow \mathrm{h}=1 \quad \therefore \mathrm{k}=-1
\]
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