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Question -

Find the distance of the point of intersection of the lines 2x + 3y = 21 and 3x – 4y + 11 = 0 from the line 8x + 6y + 5 = 0.



Answer -

Given:

The lines 2x + 3y = 21and 3x – 4y + 11 = 0

Solving the lines 2x +3y = 21 and 3x − 4y + 11 = 0 we get:

x = 3, y = 5

So, the point ofintersection of 2x + 3y = 21 and 3x − 4y + 11 = 0 is (3, 5).

Now, the perpendiculardistance d of the line 8x + 6y + 5 = 0 from the point (3, 5) is

The distance is59/10.

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