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 Post subject: GMAT Coordinate GeometryPosted: Mon Apr 29, 2013 12:52 pm

Joined: Sun May 30, 2010 2:23 am
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The two perpendicular lines L and R intersect at the point (3, 4) on the coordinate plane to form a triangle whose other two vertices rest on the x-axis. If one of the other two vertices is located at the origin, what is the x-coordinate of the other vertex?

A. 11/8
B. 33/8
C. 5
D. 20/3
E. 25/3

(E) The best way to begin solving this problem is to sketch the triangle. Since the lines are perpendicular, the triangle must be a right triangle whose right angle is at the point (3, 4).

We know that two vertices of the triangle are (0; 0) and (3; 4). The third vertex is (w; 0) because it lies on the x-axis.

There are two way to solve this problem.

I. Using equations of the lines.

The key to this problem is to find the equations of the lines that intersect and then to make some substitutions.

Since we are given two collinear points, the origin (0, 0) and (3, 4), it should be clear that the slope of the line that connects these points is equal to 4/3. So, the slope of the perpendicular line must be the negative reciprocal of that slope: -3/4.

Next, we want to find the equation of the line that passes through the point (3, 4) and has the slope -3/4. Using the point slope formula for a line, y = m(x - x1) + y1, where m is the slope, and (x1, y1) = (3, 4), we have: y = (-3/4)(x - 3) + 4 or y = (-3/4)x + (25/4). Since we know that the other vertex rests on the x-axis, where y = 0, we simply substitute y = 0 and solve for x in the equation:
0 = (-3/4)x + (25/4) or x = 25/3. We can simplify the ratio 25/3 or answer choice (E).

II. Using Pythagorean theorem.

Note, that w is positive because otherwise angle at vertex (3; 4) could not be 90 degrees.

There is a formula for finding distance between any two points on a plane.
Suppose we have two points (a, b) and (c, d). The distance between them is √[(ac)² + (bd)²]

We can use this formula for finding distances between any two vertices of the triangle:
√[(3 – 0)² + (4 – 0)²] = √[9 + 16] = √25 = 5 is the distance between (0; 0) and (3; 4).
√[(w – 0)² + (0 – 0)²] = w (because w is positive) is the distance between (0; 0) and (w; 0)
√[(w – 3)² + (0 – 4)²] = √[(w – 3)² + 16] is the distance between (3; 4) and (w; 0)

Then we use Pythagorean theorem:
w² = 5² + (√[(w – 3)² + 16])²
w² = 25 + (w – 3)² + 16
w² = 41 + w² – 6w + 9
6w = 50
w = 25/3

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