Slope of a line passing through $\mathrm{P}(2,3)$ and intersecting the line, $\mathrm{x}+\mathrm{y}=7$ at a distance of 4 units from $P$, is:

(D) $\frac{1-\sqrt{7}}{1+\sqrt{7}}$

(A) $\frac{\sqrt{5}-1}{\sqrt{5}+1}$

(B) $\frac{1-\sqrt{5}}{1+\sqrt{5}}$

(C) $\frac{\sqrt{7}-1}{\sqrt{7}+1}$

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Straight Lines

Slope of a line passing through

P (2, 3)

and intersecting the line,

x + y = 7

at a distance of 4 units from

P

, is:

(A) $\frac{5 - 1}{5 + 1}$
(B) $\frac{1 - 5}{1 + 5}$
(C) $\frac{7 - 1}{7 + 1}$
(D) $\frac{1 - 7}{1 + 7}$ ✓ Correct

Answer: (D)

Solution

Put the line through

P (2, 3)

in the form

x = 2 + r cos θ, y = 3 + r sin θ

, and let it meet

x + y = 7

at distance

r = 4

. Then

(2 + 4 cos θ) + (3 + 4 sin θ) = 7

gives

cos θ + sin θ = \frac{1}{2}

. Squaring,

1 + sin 2 θ = \frac{1}{4}

, so

sin 2 θ = - \frac{3}{4}

. With

sin 2 θ = \frac{2 m}{1 + m ^{2}}

this is

\frac{2 m}{1 + m ^{2}} = - \frac{3}{4}

, i.e.

3 m^{2} + 8 m + 3 = 0

, whose roots are

m = \frac{- 4 \pm 7}{3}

. Rationalising,

\frac{1 - 7}{1 + 7} = \frac{- 4 + 7}{3}

, which is one of these slopes.

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Single correct option

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