(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.
