Find the equation of the straight line with a positive gradient which

1.	Find the equation of the straight line with a positive gradient which passes through the point (–5, 0) and is at a perpendicular distance of 3 units from the origin. (a) 3x + 4y – 15 = 0 (b) 4x – 3y + 15 = 0 (c) 3x – 4y + 15 = 0 (d) 3y – x + 10 = 0
Answer» (c) 3x – 4y + 15 = 0 Let (m > 0) be the gradient (slope) of the required line. Then, Equation of any line through (–5, 0) having slope = m is y – 0 = m(x – (–5)) or mx – y + 5m = 0 ...(i) Its perpendicular distance from origin is 3 ⇒ \(\frac{\pm\|\,m.\,0-0+5m\,\|}{\sqrt{m^2+(-1)^2}}\) = 3 ⇒ \| 5m \| = 3\(\sqrt{m^2+1}\) \(\bigg(\because\text{Distance of}\,(x_1,y_1)\,\text{from line}\,as+by+c=\frac{\|as_1+by_1+c\|}{\sqrt{a^2+b^2}}\bigg)\) ⇒ 25m² = 9(m² + 1) ⇒ 16m² = 9 ⇒ m = \(\frac{3}{4}\) (∵ m is +ve) ∴ Required equation: y = \(\frac{3}{4}\) (x + 5) ⇒ 3x – 4y + 15 = 0.

Find the equation of the straight line with a positive gradient which passes through the point (–5, 0) and is at a perpendicular distance of 3 units from the origin. (a) 3x + 4y – 15 = 0 (b) 4x – 3y + 15 = 0 (c) 3x – 4y + 15 = 0 (d) 3y – x + 10 = 0

Answer»

(c) 3x – 4y + 15 = 0

Let (m > 0) be the gradient (slope) of the required line. Then,

Equation of any line through (–5, 0) having slope = m is

y – 0 = m(x – (–5)) or mx – y + 5m = 0 ...(i)

Its perpendicular distance from origin is 3

⇒ \(\frac{\pm|\,m.\,0-0+5m\,|}{\sqrt{m^2+(-1)^2}}\) = 3 ⇒ | 5m | = 3\(\sqrt{m^2+1}\)

\(\bigg(\because\text{Distance of}\,(x_1,y_1)\,\text{from line}\,as+by+c=\frac{|as_1+by_1+c|}{\sqrt{a^2+b^2}}\bigg)\)

⇒ 25m² = 9(m² + 1) ⇒ 16m² = 9 ⇒ m = \(\frac{3}{4}\) (∵ m is +ve)

∴ Required equation: y = \(\frac{3}{4}\) (x + 5)

⇒ 3x – 4y + 15 = 0.

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