JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Straight Lines

The equation of one of the straight lines which passes through the point

(1, 3)

and makes an angle

tan^{- 1} (2)

with the straight line,

y + 1 = 32 x

is :

(A) $42 x - 5 y - (5 + 42) = 0$
(B) $52 x + 4 y - (15 + 42) = 0$
(C) $42 x + 5 y - 42 = 0$
(D) $42 x + 5 y - (15 + 42) = 0$ ✓ Correct

Answer: (D)

Solution

The given line has slope

32

, and the required line of slope

m

meets it at an angle whose tangent is

2

2 = \frac{m - 3 2}{1 + 3 2 m}

Removing the modulus with the positive sign gives

m - 32 = 2 + 6 m

, so

m = - \frac{4 2}{5}

; with the negative sign it gives

m - 32 = - 2 - 6 m

, so

m = \frac{2 2}{7}

. Through

(1, 3)

the first slope gives

y - 3 = - \frac{4 2}{5} (x - 1) ⟹ 5 y - 15 = - 42 x + 42

so one such line is

42 x + 5 y - (15 + 42) = 0

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Single correct option

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