Find the equation of a straight line passing through the intersection of 2x+5y-4=0 with X axis and perpendicular to the line 3x-7y+8=0

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Solution

Let m be the slope of the required line
If the line is parallel to 3x-7y+8=0 then their slopes must be equal. So the slope of 3x-7y+8=0 is 3/7. So m=3/7.
Now 2x + 5y - 4 = 0 intersects the x axis at (2,0). [put y=0 in 2x + 5y - 4 = 0].
Using slope point form
y-0=3/7(x-2)
7y=3x-6
3x-7y-6=0 is the required line

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