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

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

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16 May 2016, 20:06
2
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45% (medium)

Question Stats:

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

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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

*An answer will be posted in 2 days.

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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" ##### Most Helpful Expert Reply Math Expert 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] ### Show Tags 16 May 2016, 22:25 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 _________________ ##### General Discussion Math Revolution GMAT Instructor 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] ### Show Tags 18 May 2016, 15:59 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. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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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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27 Sep 2017, 17:03
1
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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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..."
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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

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

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