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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
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GMAT 1: 760 Q51 V42 GPA: 3.82
In the x-y plane, a triangle ABC connected by 3 points A(0,4), B(-5,0)  [#permalink]

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Difficulty:   45% (medium)

Question Stats: 68% (02:42) correct 32% (02:43) wrong based on 105 sessions

### HideShow timer Statistics 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.

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Math Expert V
Joined: 02 Aug 2009
Posts: 7764
Re: In the x-y plane, a triangle ABC connected by 3 points A(0,4), B(-5,0)  [#permalink]

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3
4
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
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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 7603
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: In the x-y plane, a triangle ABC connected by 3 points A(0,4), B(-5,0)  [#permalink]

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

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

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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..."
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Posts: 3076
In the x-y plane, a triangle ABC connected by 3 points A(0,4), B(-5,0)  [#permalink]

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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

Attachment: ttttttt.png [ 10.22 KiB | Viewed 1000 times ]

gps5441 wrote
Quote:
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 $$y = 0$$

$$0 = \frac{-5}{4}x + 4$$

$$\frac{5}{4}x = 4$$

$$x = \frac{16}{5} = c$$

The value of $$c$$ is $$\frac{16}{5}$$

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

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_________________ Re: In the x-y plane, a triangle ABC connected by 3 points A(0,4), B(-5,0)   [#permalink] 05 Jan 2019, 05:39
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