JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Straight Lines

Let the area of the triangle formed by a straight line

L : x + b y + 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

4 5^{\circ}

with the positive

x

-axis, then the value of

b^{2} + c^{2}

is:

(A) $90$
(B) $93$
(C) $97$ ✓ Correct
(D) $83$

Answer: (C)

Solution

The perpendicular from the origin to the line makes

4 5^{\circ}

with the positive

x

-axis, so its slope is

1

. Being perpendicular to

L : x + b y + c = 0

, whose slope is

- \frac{1}{b}

, the product of slopes is

- 1

, which forces

b = 1

. The line

x + y + c = 0

meets the axes at

(- c, 0)

and

(0, - c)

, so the triangle it bounds with the axes has area

\frac{1}{2} c^{2} = 48

, giving

c^{2} = 96

. Hence

b^{2} + c^{2} = 1 + 96 = 97

JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Single correct option

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