Q. 124.6( 8 Votes )

# A line is such that its segment between the straight line 5x – y – 4 = 0 and 3x + 4y – 4 = 0 is bisected at the point (1, 5). Obtain its equation.

Assuming:

P1 P2 be the intercept between the lines 5x − y − 4 = 0 and 3x + 4y − 4 = 0.

P1 P2 makes an angle θ with the positive x–axis.

Given: (x1, y1) = A (1, 5)

Explanation:

So, the equation of the line passing through A (1, 5) is

Formula Used:

Let AP1 = AP2 = r

Then, the coordinates of P1 and P2 are given by

and

So, the coordinates of P1 and P2 are 1 + rcosθ, 5 + rsinθ and 1 – rcosθ, 5 – rsinθ, respectively.

Clearly, P1 and P2 lie on 5x − y − 4 = 0 and 3x + 4y − 4 = 0, respectively.

5(1 + rcos θ) – 5 – rsin θ – 4 = 0 and 3(1 – r cos θ) + 4(5 – rsin θ) + 4(5 – r sin θ) – 4 = 0

and

95 cos θ – 19 sinθ = 12 cos + 16 sinθ

83 cosθ = 35 sinθ

tan θ = 83/35

Thus, the equation of the required line is

83x – 35y + 92 = 0

Hence, the equation of line is 83x – 35 y + 92 = 0

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