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Question 271064: what is the distance from (-4,5) to the line defined by y=-2x+7
Answer by dabanfield(803)   (Show Source):
You can put this solution on YOUR website! what is the distance from (-4,5) to the line defined by y=-2x+7
We need first to find the line that passes through (-4,5) and is perpendicular to the line y = -2x+7.
We know the slope of this new line is the negative reciprocal of the slope, -2, of the line above. That slope is -(1/-2) = 1/2. The equation of the line in slope-intercept form is then;
y = (1/2)*x + b
Since the line contains the point (-4,5) we know when y = 5, x = -4 so:
5 = (1/2)*-4 + b
5 = -2 + b
b = 7
The equation of the perpendicular line is then y = (1/2)*x + 7
We need to solve the two line equations simultaneously to find the point of intersection
1.) y = -2*x + 7
2.) y = (1/2)*x + 7
Substituting -2x + 7 for y in equation 2.) we have
-2*x + 7 = (1/2)*x +7
-2*x - (1/2)*x = 0
-2.5*x = 0
x = 0
From 1.) we have y = -2*x +7 = -2*0 + 7 = 7
So the point of interesction is (0,7)
The distance then from (-4,5) to (0,7) is:
sqrt[(7-5)^2 + (0-(-4))^2] =
sqrt(2^2 + 4^2) =
sqrt(20) = sqrt(4*5) = sqrt(4)*sqrt(5) = 2*sqrt(5)
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