Question : $P(4,2)$ and $R(–2,0)$ are vertices of a rhombus $PQRS$. What is the equation of diagonal $QS$?
Option 1: $x–3y=–2$
Option 2: $3x+y=4$
Option 3: $3x+y=–4$
Option 4: $x–3y=2$
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Correct Answer: $3x+y=4$
Solution :
$QS$ perpendicularly bisects the diagonal joining $P(4,2)$ and $R(–2,0)$ at $O$, thus $O$ is the mid-point of $PR$ and $QS$.
Co-ordinates of point $O$ = $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$
= $(\frac{4–2}{2},\frac{2+0}{2})=(1,1)$
The slope of the straight line $PR$
= $\frac{y_2–y_1}{x_2–x_1}=\frac{0–2}{–2–4}$
= $\frac{–2}{–6}=\frac{1}{3}$
We know that $m_1m_2=–1$ $(\because PR\perp QS)$
Therefore, slope of $QS$ = $–\frac{1}{\frac{1}{3}}=–3$
Equation of the straight line $QS$ passing through the point (1, 1)
⇒ $y–1=–3(x–1)$
⇒ $y–1=–3x+3$
⇒ $3x+y=4$
Hence, the correct answer is $3x+y=4$.
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