JEE Mains · Maths · STD 11 - 9. straight line
A rectangle is formed by the lines \( x=0, x=3, y=0 \) and \( y=4 \). Let the line L be perpendicular to \( 3x+y+6=0 \) and divide the area of the rectangle into two equal parts. Then the distance of the point \( (\frac{1}{2},-5) \) from the line L is equal to:
- A \( 2\sqrt{5} \)
- B \( 3\sqrt{10} \)
- C \( \sqrt{10} \)
- D \( 2\sqrt{10} \)
Answer & Solution
Correct Answer
(D) \( 2\sqrt{10} \)
Step-by-step Solution
Detailed explanation
Line is \(y=\frac{x}{3}+C\) Line passes thru \(\left(\frac{3}{2}, 2\right)\) \(2=\frac{1}{2}+C \Rightarrow C=\frac{3}{2}\) \(y=\frac{x}{3}+\frac{3}{2}\) \(\Rightarrow 6 y=2 x+9\) Line is \(2 x-6 y+9=0\) & Dist \(=\left|\frac{1+30+9}{\sqrt{40}}\right|=\sqrt{40}=2 \sqrt{10}\)
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