Question 38647
Start by putting the given equation in the slope-intercept form: {{{y = mx + b}}}

{{{2x-5y = 7}}} --> {{{y = (2/5)x - 7/5}}} You can see that the slope, m, of the given line is {{{2/5}}}
Now you want the family of lines that are perpendicular to this line, and as you know, perpendicular lines will have a slope that is the negative reciprocal of the given line...so the slope of the perpendicular lines will be the negative reciprocal of {{{m = 2/5}}} or {{{m = -(5/2)}}}
Since you are looking for the family of lines, you don't really care where these lines intercept the y-axis, so the y-intercept, b, can be any real number.
So now we have, in the slope-intercept form, the equation for the family of lines perpendicular to the given line:
{{{y = -(5/2)x + b}}} Where b, the y-intercept is any real number. Now if you rearrange this equation:
{{{y = -(5/2)x + b}}} Multiply both sides by 2 to clear the fraction.
{{{2y = -5x + 2b}}} Add 5x to both sides.
{{{5x + 2y = 2b}}} But since b is any real number, so 2b is any real number so you can let 2b = k, so finally you have:
{{{5x + 2y = k}}}.