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by yumi2012 » Sat Aug 24, 2013 11:26 pm
What are the coordinates for the point on Line AB (see figure) that is three times as far from A as from B, and that is in between points A and B?
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by ganeshrkamath » Sun Aug 25, 2013 8:21 am
yumi2012 wrote:What are the coordinates for the point on Line AB (see figure) that is three times as far from A as from B, and that is in between points A and B?
Image
Equation of the line is y/(x+2) = 6/(-5+2)
=> y/(x+2) = -2____________________(1)

Distance between A and B = sqrt((-5-(-2))^2 + (6-0)^2) = sqrt(9+36) = sqrt(45)
Distance between new point C and B = sqrt(45)/4
sqrt(45)/4 = sqrt((x+2)^2 + y^2)
45/16 = (x+2)^2 + y^2

From (1), (x+2) = -y/2
45/16 = (-y/2)^2 + y^2
45/16 = y^2/4 + y^2
45/16 = 5y^2/4
y^2 = 9/4
y = 3/2
x = -(3/2)/2 - 2
x = -3/4 - 2
x = -11/4

So the coordinates of the point are [spoiler](-11/4, 3/2)[/spoiler]

Cheers

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by Brent@GMATPrepNow » Sun Aug 25, 2013 8:49 am
yumi2012 wrote:What are the coordinates for the point on Line AB (see figure) that is three times as far from A as from B, and that is in between points A and B?
Image
Here's a different approach.
Consider the right triangle with AB as the hypotenuse.
Image

Now divide the two legs into 4 parts.
Image

If we draw a right triangle like so, we'll see that the red triangle is 1/4 the size of the original blue triangle.
Image

IMPORTANT: This means that the distance from A to C is THREE TIMES the distance from C to B. So, point C satisfies the given information.

Now all we need to do is determine the size of the red triangle.
Image

Since the red triangle is 1/4 the size of the blue triangle, its sides have lengths 3/4 and 1 1/2.
So, to the find the coordinates of C, just factor these lengths into the coordinates of point B.

So, the x-coordinate of point C = (-2) - (3/4) = -2 3/4 (aka -2.75)
The y-coordinate of point C = (0) + (1 1/2) = 1 1/2 (aka 1.5)

So the coordinates are [spoiler](-2.75, 1.5)[/spoiler]

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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