Question

In what direction should a line be drawn through the point
(1,2), so that its point of intersection with the line x+y=4
is at a distance root 6/3 from the given point.​

Answer 1

$\Huge{\underline{\underline{\mathfrak{ Solution \colon }}}}$

Equation of a line passing through (x1,y1) and making an angle θ is x−x1cosθ=y−y1sinθ=r where r is the distance of the point from the given line.

$\bold \: \underline\mathfrak{Step \: by \: step \: explanation:}$

Step 1:

Let the required line make an angle θ with x - axis.

$:⟹\frac{\text{x - 1}}{ \cosθ } = \frac{ \text{y - 2}}{ \sin θ } = \frac{ \sqrt{6} }{3}$

$: ⟹ \text{x} = \frac{ \sqrt{6 } \cos θ }{ 3} + 1$

$:⟹ \text{y} = \frac{ \sqrt{6} \sin θ}{3} + 2$

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Step 2 :

This points lies on the line x + y = 4

$: ⟹ \frac{ \sqrt{6 } \cos θ }{ 3} + 1 + \frac{ \sqrt{6} \sinθ}{3} + 2 = 4$

$: ⟹ \frac{ \sqrt{6 } \cos θ }{ 3} + \frac{ \sqrt{6} \sinθ}{3} = 1$

$: ⟹ \sqrt{6} \ \cos(?) θ + \sqrt{6} \sinθ$

$: ⟹ \cosθ + \sinθ = \frac{3}{ \sqrt{6} }$

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Step 3 :

Squaring on both sides

$: ⟹ {(cos θ + \sinθ) }^{2} = \frac{3}{2}$

$\text{But sin }\: 2θ+cos2θ=1$

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∴2sinθcosθ=32−1

⇒ 2sinθcosθ=sin2θ

∴sin2θ=12

⇒ 2θ=π6 or 30∘

∴θ=15∘ or 90∘−15∘=75∘

Hence the required line is in the

positive direction of either 15∘ or 75∘ to the x axis.

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