What is the equation of the line through the point (3, 4) and perpendicular to 7x + 5y – 10 =0?

If a line is said to be perpendicular to a given line, its slope
should be equal to the negative reciprocal of the given line.

The
given line is 7x + 5y - 10 = 0.  To get the slope of this given equation, we have to transform
this equation in the slope-intercept form y = mx +b.  The equation
becomes,

5y = -7x + 10

y = (-7/5)x +
2

so the slope is -7/5.

So the slope of
the new line perpendicular to the given line is 5/7 since it should be negative reciprocal, and
it should pass the point (3,4)

Using point-slope form y - y1 = m(x -
x1)

y - 4 = (5/7)(x - 3)

Multiply the
whole equation by 7 it becomes,

7y - 28 = 5(x -
3)

7y - 28 = 5x - 15

So transposing all
the terms in one side, the equation becomes

5x - 7y
+13 = 0.