A) \[x-3y=-2\]
B) \[3x+y=4\]
C) \[3x+y=-4\]
D) \[x-3y\text{=}2\]
Correct Answer: B
Solution :
(b): The diagonals of a rhombus bisect each other at right angles. \[\therefore \]Co-ordinates of point 'O' \[=\left( \frac{{{x}_{1}}+{{x}_{2}}}{2},\frac{{{y}_{1}}+{{y}_{2}}}{2} \right)\] \[=\left( \frac{4-2}{2},\frac{2+0}{2} \right)=\left( 1,1 \right)\] Slope of straight line PR \[=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{2}}}=\frac{0-2}{-2-4}\] \[=\frac{-2}{-6}=\frac{1}{3}\] \[\therefore \] \[PR\bot OS\] \[\therefore \] Slope of \[QS=-\frac{1}{\frac{1}{3}}=-3\] \[[\therefore {{m}_{1}}{{m}_{2}}=-1]\] \[\therefore \]Equation of straight line QS passing through point (1.1): \[y-1=-3\left( x-1 \right)\] \[\Rightarrow \]\[y-1=3x+3\] \[\Rightarrow \]\[3x+y=4\]
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