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Find the equation of a straight line which makes acute angle with positive direction of x-axis passes through point (– 5, 0) and is at a perpendicular distance of 3 units from the origin.

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Let ‘a’ be the acute angle made by the line with positive x-axis.

∴ Equation of the line is x cos a + y sin a = 3 …(1)

Since the line passes through (– 5, 0), so

⇒ – 5 cos a + 0. Sin a = 3

⇒ cos a = – \(\frac{3}{5}\)

⇒ sin a = \(\sqrt{1 − (-\frac{3}{5})^2}\) = \(\sqrt{1 − \frac{9}{25}}\) = \(\frac{4}{5}\)

(1) ⇒ x(− \(\frac{3}{5}\)) + y(\(\frac{4}{3}\)) = 3

⇒ 3x – 4y + 15 = 0, is the required of line.

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