Slope of a line passing through and intersecting the line, at a distance of 4 units from , is
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: C
Solution
Using parametric form, any point at distance 4 from P(2,3) along a line making angle θ with x-axis is (2 + 4cosθ, 3 + 4sinθ). This point lies on x + y = 7, so:
(2 + 4cosθ) + (3 + 4sinθ) = 7
4cosθ + 4sinθ = 2
cosθ + sinθ = 1/2
Let slope m = tanθ = sinθ/cosθ. From cosθ + sinθ = 1/2, square both sides:
1 + 2sinθcosθ = 1/4
sin2θ = -3/4
Also, (cosθ - sinθ)² = 1 - sin2θ = 1 + 3/4 = 7/4, so cosθ - sinθ = ±√7/2.
Combined with cosθ + sinθ = 1/2:
- Case 1: cosθ = (1 + √7)/4, sinθ = (1 - √7)/4 → tanθ = (1 - √7)/(1 + √7)
- Case 2: cosθ = (1 - √7)/4, sinθ = (1 + √7)/4 → tanθ = (1 + √7)/(1 - √7)
The slope (1 - √7)/(1 + √7) matches option 3.
(2 + 4cosθ) + (3 + 4sinθ) = 7
4cosθ + 4sinθ = 2
cosθ + sinθ = 1/2
Let slope m = tanθ = sinθ/cosθ. From cosθ + sinθ = 1/2, square both sides:
1 + 2sinθcosθ = 1/4
sin2θ = -3/4
Also, (cosθ - sinθ)² = 1 - sin2θ = 1 + 3/4 = 7/4, so cosθ - sinθ = ±√7/2.
Combined with cosθ + sinθ = 1/2:
- Case 1: cosθ = (1 + √7)/4, sinθ = (1 - √7)/4 → tanθ = (1 - √7)/(1 + √7)
- Case 2: cosθ = (1 - √7)/4, sinθ = (1 + √7)/4 → tanθ = (1 + √7)/(1 - √7)
The slope (1 - √7)/(1 + √7) matches option 3.
[JEE Main 2019 · 9 April · Shift I]