Let the area of the triangle formed by a straight line L : x + by + c = 0 with co-ordinate axes be 48 square units. If the perpendicular drawn from the origin to the line L makes an angle of 45° with the positive x-axis, then the value of is : [

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.
Let the area of the triangle formed by a straight line L : x + by + c = 0 with co-ordinate axes be 48 square units. If the perpendicular drawn from the origin to the line L makes an angle of 45° with the positive x-axis, then the value of $b^{2} + c^{2}$ is : [2025]

1 83

2 97

3 93

4 90

Ans.
(2)

We have, area of $△$ OAB = 48 sq. units

Now, slope of line OD is m = tan 45° = 1

$∴$ Slope of line $A B = \frac{- 1}{m} = - 1$ ... (i)

Also, equation of line AB is x + by + c = 0

So, Slope $\frac{- 1}{b}$ ... (ii)

From (i) and (ii), we get

$\Rightarrow - 1 = \frac{- 1}{b} \Rightarrow b = 1$

Now, x and y intercept of line x + by + c = 0 are given as x = – c and $y = \frac{- c}{b}$ respectively.

$∴$ Area of $△$ OAB = $\frac{1}{2} \times (- c) \times \frac{- c}{b} \Rightarrow 48 = \frac{c^{2}}{2 b}$

$\Rightarrow c^{2} = 48 \times 2$ [ $∵$ b = 1]

$\Rightarrow c^{2} = 96$

So, $b^{2} + c^{2} = 1 + 96 = 97$ .

Want Us to Email you About Special Offers & Updates?