+3 votes
in JEE by kratos

Equation of a straight line passing through the point (4, 5) and equally inclined to the lines 3x = 4y + 7 and 5y = 12x + 6 is (angle bisector)

(a) 9x – 7y = 1

(b) 9x + 7y = 71

(c) 7x – y = 73

(d) 7x – 9y + 17 = 0

1 Answer

+2 votes
by kratos
 
Best answer

The correct option (a) 9x – 7y = 1

Explanation:

Line is equally inclined to lines 3x – 4y – 7 = 0 and 12x – 5y + 6 = 0

∴ [(3x – 4y – 7)/√(32 + 42)] = ± [(12x – 5y + 6)/√{122 + 52}]

∴ [(3x – 4y – 7)/5] = ± [(12x – 5y + 6)/(13)]

i.e. 21x + 27y + 121 = 0

99x – 77y – 61 = 0

i.e. their slopes are [(– 7)/9] and (9/7)

∴ equation of line with slope [(– 7)/9] and through point (4, 5) is

y – 5 = [(– 7)/9](x – 4) ⇒ 7x + 9y = 73

equation of line with slope (9/7) and through point (4, 5) is

y – 5 = (9/7)(x – 4) ⇒ 9x – 7y = 1

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