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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

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.

Hi, GOOD Q..

we can know from the information that AB is PERPENDICULAR to AC... so BEST method would be to find the slope.. slope of AB = \(\frac{0-4}{-5-0} =\frac{4}{5}\) so slope of perpendicular, AC, will be \(\frac{-5}{4}\)... also as per coord of A and C, the slope is \(\frac{0-4}{c-0}= \frac{-4}{c} = \frac{-5}{4}.......................c = \frac{16}{5}\)............. A
_________________

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.
_________________

Re: In the x-y plane, a triangle ABC connected by 3 points A(0,4), B(-5,0) [#permalink]

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27 Sep 2017, 17:03

1

This post was BOOKMARKED

MathRevolution wrote:

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.

I understand how to do this problem. First, I plotted the points for visual (I understand this is not needed.) Then I found the slope of A to B. and then I did trial and error to find a answer with the negative reciprocol. Luckily the answer was A so it took me about 2 mins but if the answer was say D or E, this problem would take a lot longer. Is there any tricks to solve this quicker?

Re: In the x-y plane, a triangle ABC connected by 3 points A(0,4), B(-5,0) [#permalink]

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18 Oct 2017, 05:10

chetan2u wrote:

MathRevolution 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.

Hi, GOOD Q..

we can know from the information that AB is PERPENDICULAR to AC... so BEST method would be to find the slope.. slope of AB = \(\frac{0-4}{-5-0} =\frac{4}{5}\) so slope of perpendicular, AC, will be \(\frac{-5}{4}\)... also as per coord of A and C, the slope is \(\frac{0-4}{c-0}= \frac{-4}{c} = \frac{-5}{4}.......................c = \frac{16}{5}\)............. A

Can you please explain how you got this information. I tried to plot the points but unable to get AB perpendicular to AC "we can know from the information that AB is PERPENDICULAR to AC..."

In the x-y plane, a triangle ABC connected by 3 points A(0,4), B(-5,0) [#permalink]

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18 Oct 2017, 13:12

MathRevolution 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

Can you please explain how you got this information. I tried to plot the points but unable to get AB perpendicular to AC "we can know from the information that AB is PERPENDICULAR to AC..."

I had a hard time with this for a bit, too.

This

Quote:

If the angle from a point A is 90o

Means this

Quote:

If the angle from point A is 90°

If there is a right angle at point A (see diagram), then AB is perpendicular to AC. (I just freehand sketched a line to get a visual once I figured out how to read the question; as you noted, it does not have to be done . . . but it helps.)

And if perpendicular, line AC will have a slope that is the negative reciprocal of the slope for AB.

So AB's slope is \(\frac{rise}{run} = \frac{(0 - 4)}{(-5 - 0)} = \frac{-4}{-5} = \frac{4}{5}\)

Take the negative reciprocal; AC's slope will be \(\frac{-5}{4}\)

Equation for the line of which AC is a segment (I used this method, but "product of slopes = -1" is probably faster): \(y = mx + b\)

Slope = \(\frac{-5}{4}\), y-intercept (from point A) = \(4\)

\(y = -\frac{5}{4}x + 4\)

The \(x\)-intercept will yield \(c\) in point C's (\(c,0\)); set \(y = 0\)