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.

Answer :

**Assuming:**

P_{1} P_{2} be the intercept between the lines 5x − y − 4 = 0 and 3x + 4y − 4 = 0.

P_{1} P_{2} makes an angle θ with the positive x–axis.

**Given:** (x_{1}, y_{1}) = A (1, 5)

**Explanation:**

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

**Formula Used:**

Let AP_{1} = AP_{2} = r

Then, the coordinates of P1 and P2 are given by

and

So, the coordinates of P_{1} and P_{2} are 1 + rcosθ, 5 + rsinθ and 1 – rcosθ, 5 – rsinθ, respectively.

Clearly, P_{1} and P_{2} 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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