Find the equation of the line perpendicular to x−5y=15 that passes through the point (−2,5).
Question
2 Answer

1. User Answers harpazo
1. Solve the given equation for y.
x  5y = 15
5y = x + 15
y = (x + 15)/5
y = (x/5)  3
y = (1/5)(x)  3
The slope is 1/5. See it?
The equation we are looking for has a slope which is the negative inverse of the slope in the equation we just solved for y.
The slope for the equation we want is 5 which is the negative inverse of 1/5. Undetstand?
We have the slope of the new equation and one point is given.
Plot BOTH into the pointslope formula and solve for y. To solve for a variable means to isolate the variable ALONE on one side of the equation.
y  y_1 = m(x  x_1)...This is the pointslope formula. Our given point is (5,2)
y  5 = 5(x  (2))
y  5 = 5(x + 2)
We now solve for y and that's it.
y  5 = 5x  10
y = 5x  10 + 5
The equation we want is y = 5x  5. 
2. User Answers hem3587
1. Solve the given equation for y.
x  5y = 15
5y = x + 15
y = (x + 15)/5
y = (x/5)  3
y = (1/5)(x)  3
The slope is 1/5. See it?
The equation we are looking for has a slope which is the negative inverse of the slope in the equation we just solved for y.
The slope for the equation we want is 5 which is the negative inverse of 1/5. Undetstand?
We have the slope of the new equation and one point is given.
Plot BOTH into the pointslope formula and solve for y. To solve for a variable means to isolate the variable ALONE on one side of the equation.
y  y_1 = m(x  x_1)...This is the pointslope formula. Our given point is (5,2)
y  5 = 5(x  (2))
y  5 = 5(x + 2)
We now solve for y and that's it.
y  5 = 5x  10
y = 5x  10 + 5
The equation we want is y = 5x  5.
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