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In the x-y plane, a triangle ABC connected by 3 points A(0,4

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In the x-y plane, a triangle ABC connected by 3 points A(0,4

by Max@Math Revolution » Mon May 16, 2016 7:06 pm
In the x-y plane, a triangle ABC connected by 3 points A(0,4), B(-5,0), and C(c,0) is there. If the angle from a point A is 90o, what is the value of c?
A. 16/5
B. 12/5
C. 9/5
D. 7/5
E. 3/5

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by 800_or_bust » Tue May 17, 2016 6:03 am
Max@Math Revolution wrote:In the x-y plane, a triangle ABC connected by 3 points A(0,4), B(-5,0), and C(c,0) is there. If the angle from a point A is 90o, what is the value of c?
A. 16/5
B. 12/5
C. 9/5
D. 7/5
E. 3/5

*An answer will be posted in 2 days.
The line segment connecting point B with point A has a slope of 4/5: (y2 - y1) / (x2 - x1) = (4 - 0) / (0 - (-5)) = 4/5.

Since we are told that Angle A is a right angle, this means that the line segment connecting points B and C must be perpendicular to the line segment connecting points A and B. To be perpendicular, the slope of this second line must be the negative reciprocal of line segment AB. Thus, line segment BC has slope of -5/4.

Armed with this knowledge, and the fact that the y-coordinate of point C is 0, we can simply set up a ratio and solve for x.

We know that from point A to point C we go down four units, thus -5/4 = -4/x.

Solving for x: x = 16/5. Answer choice A.
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by Max@Math Revolution » Wed May 18, 2016 3:01 pm
The slope of the line that passes through the points A and B is =(4-0)/(0-(-5))=4/5.
The slope of the line that passes through the points A and C is (4-0)/(0-c)=-4/c.
As they meet perpendicular, the product of the slope is -1. So, (4/5)(-4/c)=-1. So, we get 16/5c=1, c=16/5. Hence, the correct answer is A.
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