What is the equation of the line that passes through the point (4,-5) and is perpendicular to 2x-5y= -10?

1 Answer
KillerBunny
Oct 20, 2015

#y=-5/2x+5#

Explanation:

Rewrite the equation of the line we must be perpendicular to as #y=(2x+10)/5= 2/5 x + 2#. This is the slope-intercept form, and indeed we can see that the slope is #m=2/5#, and the intercept is #q=2# (even if we don't care about it in this specific case).

A line with slope #n# is perpendicular to a line with slope #m# if and only if the following equation holds:

#n=-1/m#.

In our case, the slope must be #-1/(2/5)=-5/2#.

So, now we know everything we need, since the slope and a known point identify a line uniquely: we can find the equation with the formula

#y-y_0=m(x-x_0)#, if #m# is the slope of the line and #(x_0,y_0)# is the known point. Plugging the values, we have

#y+5=-5/2(x-4)#, which we can adjust into

#y=-5/2x+5#

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