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Question

Find the distance of the point (2, 5) from the line 3x + y + 4 = 0 measured parallel to the line 3x − 4y + 8 = 0.

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Solution

Here, x1, y1=A2, 5
It is given that the required line is parallel to 3x −4y + 8 = 0

4y=3x+8y=34x+2

tanθ=34sinθ=35, cosθ=45

So, the equation of the line is

x-x1cosθ=y-y1sinθx-245=y-5353x-6=4y-203x-4y+14=0

Let the line 3x-4y+14=0 cut the line 3x + y + 4 = 0 at P.

Let AP = r
Then, the coordinates of P are given by

x-245=y-535=r

x=2+4r5, y=5+3r5

Thus, the coordinates of P are 2+4r5, 5+3r5.

Clearly, P lies on the line 3x + y + 4 = 0.

32+4r5+ 5+3r5+4=0 6+12r5+ 5+3r5+4=015+15r5=0r=-5

∴ AP = r = 5

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