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:


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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