A rectangle is formed by the lines , and . Let the line be perpendicular to and divide the area of the rectangle into two equal parts. Then the distance of the point from the line is equal to: [2026]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
A rectangle is formed by the lines $x = 0, y = 0, x = 3$ , and $y = 4$ . Let the line $L$ be perpendicular to $3 x + y + 6 = 0$ 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: [2026]

1 $\sqrt{10}$

2 $2 \sqrt{5}$

3 $2 \sqrt{10}$

4 $3 \sqrt{10}$

Ans.
(3)

$Line is y = \frac{x}{3} + C$

$Line passes thru (\frac{3}{2}, 2)$

$2 = \frac{1}{2} + C \Rightarrow C = \frac{3}{2}$

$y = \frac{x}{3} + \frac{3}{2}$

$\Rightarrow 6 y = 2 x + 9$

$Line is 2 x - 6 y + 9 = 0$

$Dist =$ $| \frac{1 + 30 + 9}{\sqrt{40}} |$ $= \sqrt{40} = 2 \sqrt{10}$

Want Us to Email you About Special Offers & Updates?