Find the equation of the line which passes through the point of inters

1.	Find the equation of the line which passes through the point of intersection of the lines 2x – y + 5 = 0 and 5x + 3y – 4 = 0 and is perpendicular to the line x – 3y + 21 = 0. (a) 2x + y + 10 = 0 (b) 3x + y + 21 = 0 (c) 3x + y = 0 (d) 3y – x + 21 = 0
Answer» (c) 3\(x\) + y = 0 The equations of the two lines whose point of intersection is needed are: 2\(x\) – y = –5 ...(i) 5\(x\) + 3y = 4 ...(ii) 3 x Eqn (i) + Eqn (ii) ⇒ (6\(x\) – 3y) + (5\(x\) + 3y) = –15 + 4 ⇒ 11\(x\) = –11 ⇒ \(x\) = –1 Putting \(x\) = –1 in (i), we get –2 – y = –5 ⇒ y = 3. ∴ Point of intersection is (–1, 3). Slope of line \(x\) – 3y + 21 = 0, i.e., y = \(\frac{x}{3}\) + 7 is \(\frac{1}{3}.\) ⇒ Slope of line perpendicular to line \(x\) – 3y + 21 = 0 is –3 [∵ m₁ × m₂ = –1] ∴ Equation of a line through (–1, 3) with slope –3 is y – 3 = – 3 (\(x\) + 1) ⇒ y – 3 = –3\(x\) – 3 ⇒ 3\(x\) + y = 0.

Find the equation of the line which passes through the point of intersection of the lines 2x – y + 5 = 0 and 5x + 3y – 4 = 0 and is perpendicular to the line x – 3y + 21 = 0. (a) 2x + y + 10 = 0 (b) 3x + y + 21 = 0 (c) 3x + y = 0 (d) 3y – x + 21 = 0

Answer»

(c) 3\(x\) + y = 0

The equations of the two lines whose point of intersection is needed are:

2\(x\) – y = –5 ...(i)

5\(x\) + 3y = 4 ...(ii)

3 x Eqn (i) + Eqn (ii) ⇒ (6\(x\) – 3y) + (5\(x\) + 3y) = –15 + 4

⇒ 11\(x\) = –11 ⇒ \(x\) = –1

Putting \(x\) = –1 in (i), we get –2 – y = –5 ⇒ y = 3.

∴ Point of intersection is (–1, 3).

Slope of line \(x\) – 3y + 21 = 0, i.e., y = \(\frac{x}{3}\) + 7 is \(\frac{1}{3}.\)

⇒ Slope of line perpendicular to line \(x\) – 3y + 21 = 0 is –3

[∵ m₁ × m₂ = –1]

∴ Equation of a line through (–1, 3) with slope –3 is

y – 3 = – 3 (\(x\) + 1) ⇒ y – 3 = –3\(x\) – 3 ⇒ 3\(x\) + y = 0.

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