Question

find an equation of (-1,-2) perpendicular to 2x+5y+8=0

Answer 1

first we are going to put the equation in y = mx + b form...
2x + 5y + 8 = 0
2x + 5y = -8
5y = -2x - 8
y = -2/5x - 8/5
what we just did was find the slope of the equation. With the formula, y = mx + b, the m position is the slope, therefore the slope of this equation is -2/5.
We are looking for a perpendicular line, so the slope we need is the negative reciprocal of the slope of the equation. All that means is " flip " the fraction and change the sign, and you will have the negative reciprocal. So the negative reciprocal of - 2/5 is 5/2. That is the slope we need.

Now, since we have the slope and a set of points, we will use this formula :
y - y1 = m(x - x1)
slope(m) = 5/2
(-1,-2) x1 = -1 and y1 = -2

now lets sub our info into the formula :
y - y1 = m(x - x1)
y - (-2) = 5/2(x - (-1)
y + 2 = 5/2(x + 1)
y + 2 = 5/2x + 5/2
y = 5/2x + 5/2 - 2
y = 5/2x + 5/2 - 4/2
y = 5/2x + 1/2 <-- your equation