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13 Nov 2013, 01:20
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
64% (02:55) correct
36%
(02:49)
wrong
based on 327
sessions
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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
A. 11/8
B. 33/8
C. 5
D. 20/3
E. 25/3
13 Nov 2013, 01:56
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honchos
Look at the diagram below:
Attachment:
Untitled.png [ 6.93 KiB | Viewed 9299 times ]
Thus the slope of line L passing (0, 0) and (3, 4) is \(m=\frac{y_2-y_1}{x_2-x_1}=\frac{4-0}{3-0}=\frac{4}{3}\).
The slope of line R, which is perpendicular to L, is negative reciprocal of the slope of L, hence its \(-\frac{3}{4}\).
Now, we need to find equation of R.
The equation of a straight line that passes through a point \(P_1(x_1, y_1)\) with a slope m is: \(y-y_1=m(x-x_1)\). Therefore, the equation of R is \(y-4=-\frac{3}{4}(x-3)\) --> \(y=-\frac{3}{4}x+\frac{25}{4}\). The x-intercept of this line is 25/3.
Answer: E.
For more check here: math-coordinate-geometry-87652.html
General Discussion
ShashankDave
Joined: 03 Apr 2013
Last visit: 26 Jan 2020
Posts: 215
Given Kudos: 872
Location: India
Concentration: Marketing, Finance
Schools: McDonough '21 (A) Kelley '21 ISB '21 (I)
GMAT 1: 740 Q50 V41
GPA: 3
16 Nov 2013, 12:55
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Hi honchos...you could even do this :-
as bunuel has calculated, the slope of the line will be -3/4. Now suppose the co-ordinates of the point needed are (a,0), just equate the slopes.
You will get the equation
-3/4 = 4/(3-a)
solving this you get the OA 25/3 i.e. E.
Kudos me!
as bunuel has calculated, the slope of the line will be -3/4. Now suppose the co-ordinates of the point needed are (a,0), just equate the slopes.
You will get the equation
-3/4 = 4/(3-a)
solving this you get the OA 25/3 i.e. E.
Kudos me!
