Question

Find an equation of a line perpendicular to -2x + 5y = -10 passing through (-2,10).

Answer 1

To find this you would change this equation in standard form to y=mx+b:
-2x+5y=-10
+2x +2x
5y=2x-10 and divide both sides by 5 which gives you
y=2/5x-2 and for it to be perpendicular the slope has to be the reciprocal of that one so you flip the numerator and denominator and change from positive to negative which is -5/2 then you plug that slope into a point-slope formula which would be:
y-10=-5/2(x+2) (the negatives would cancel out and they become positive
DISTRIBUTE THE -5/2
y-10=-5/2x-5
+10 +10
y=-5/2+5
which is an equation of a line that is perpendicular

Answer 2

-2x + 5y = -10
5y = 2x - 10
y = 2/5x - 2...slope here is 2/5. A perpendicular line will have a negative reciprocal slope. All that means if flip the original slope and change the sign. So our perpendicular line will have a slope of -5/2.

y = mx + b
slope(m) = - 5/2
(-2,10)...x = -2 and y = 10
now we sub into our formula and solve for b, the y int
10 = -5/2(-2) + b
10 = 5 + b
10 - 5 = b
5 = b

so ur perpendicular equation is : y = -5/2x + 5