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20 May 2018, 21:02
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abhi758
We need 3 vertices. We can find them from the intersection of two lines.
y = 0 means that x can be anything . . . but set y = 0 and y= 5 - x equal.
The second equation "stops" y = 0 from going to the right infinitely.
Write equations for all three lines in slope-intercept form in order to keep intercepts straight.
y = mx + b
m = slope
b = y-intercept
Line 1
\(y = 5 - x\)
\(y = -x + 5\)
Line 2
\(y = 0\)
x can be any value. y is always 0
We need Line 1's x-intercept to give us a stop point
Line 3
\(2y = 3x\)
\(y = \frac{3}{2}x + 0\)
Any line that passes through the origin has slope-intercept form: \(y = mx\)
When x is 0, y is 0 and vice versa
The y-intercept is 0
The x-intercept is 0
So coordinates of one vertex are (0,0)
If that is not apparent, just find one of three vertices.
Line 2 = Line 3 will give you the same coordinates --
setting them equal means "this is the point at which the lines intersect"
Each vertex lies at the intersection point of TWO lines
To find that point, set equations equal, two at a time
We can do so because y = y
I think I see some confusion: "x" and "y" are generalized variables
we use, but the x's and y's are not all the same values.
I will use x and y to solve, but I will call
the exact counterparts of x,y coordinates
at the three different vertices (a,b), (c,d) and (e,f)
Vertex 1 (Line 1 equation = Line 3 equation):
\(-x + 5 = \frac{3}{2}x\)
\(-2x + 10 = 3x\)
\(10=5x\)
\(x=2\)
Plug THAT x-coordinate into EITHER equation to find y-coordinate of the intersection
\(y = (-x + 5) = (-2 + 5) = 3\)
Coordinates\((a,b)\) for that intersection point are \((2, 3)\)
Vertex 2 (Line 1 = Line 2) :
\(-x + 5 = 0\)
\(x = 5\)
Plug that x-value into \(y = -x + 5\)
Coordinates \((c,d)\) for that vertex are \((5,0)\)
Vertex 3 (Line 2 = Line 3):
\(\frac{3}{2}x = 0\)
\(x = 0\) (0 divided by anything except itself = 0)
Plug \(x = 0\) into \(y = \frac{3}{2}x\)
Coordinates (e,f) for that vertex are (0,0)
When two points have an x- or y-coordinate that is equal,
subtract one NON-equal coordinate from the other NON-equal coordinate.
That's length of one side of the triangle
(c,d): (5,0)
(e,f): (0,0)
The y-coordinate is 0. This side of the triangle lies on the x-axis.
\(x2 - x1 = (c - e)\) = length of a side
\((5 - 0) = 5\), which we can call the base of the triangle
The height of the triangle? Sketch if needed.
If the base lies on the x-axis, the height starts from . . . 0
Only one pair has a y-coordinate that is NOT 0
\((a,b) = (2, 3)\)
\(y2 - y1 = (b - 0) = (3 - 0) = 3\)
The triangle's height is \(3\)
Area =\(\frac{b*h}{2}\)
Area =\(\frac{5*3}{2}\)
Area = \(7.5\)
Answer A
dave13 -
Quote:
You found the intersection point for two lines.
You found coordinates for one vertex. Correct.
3 happens to be the height because the
base lies on y = 0, so the y-value, the height, is y2 - y1 = (3 - 0 ) = 3
But the base is not x=2.
No matter what y is, x always equals 2.
x = 2 is an infinite vertical line.
No side of this triangle is infinite.
What about the other two vertices?
What about the two other possible line intersections?
Plug y = 0 into y = 5 - x. Now what do you have?
It might prevent confusion if you write (x2,y2).
Hint: y2 = 0. When you plug y = 0 into that equation, you get a different x-value than the one you found above.
See my post; x and y are generalized variables but
each vertex has a specific different value of x,y
(I labeled each x,y set as a,b and c,d and e,f to clear confusion)
Plug y = 0 into 2y = 3x. Now what do you have?
It might help if you write (x3, y3)
Hope that helps!
20 May 2018, 21:54
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NandishSS
hi
you are not going wrong. you correctly got one of the vertices (0,0).
the other two are got from the following:
solving y=5-x and y=0 , we get 0=5-x so x=5. vertex is (5,0)
solving y=5-x and 2y=3x, we get 5-x=3x/2 => 10-2x=3x so x=2, y=3. vertex is (2,3)
0,0 and 5,0 lie on the X-axis. so breadth b of triangle = distance between these vertices = 5-0 =5
height of the triangle is 3( as 2,3 is the other vertex)
Area = 0.5bh= 0..5*5*3 =7.5.
option A
21 May 2018, 03:19
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dave13
Hey dave13
Without a doubt, the GMATClub tests are harder than the quant questions in the GMAT
but the important thing when we practice using these questions is to understand the
pattern of mistakes we make and work on the topics where we make mistakes often.
Also, I would recommend taking the GMATClub quants once you start hitting at least
Q40+ in the GMATPrep tests.
Now coming back to the problem and the doubt you had in it.
Attachment:
Triangle.png [ 5.4 KiB | Viewed 3061 times ]
In the above triangle, the base can be calculated using the distance formula or the two points (0,0) & (5,0).
Distance between points(base) = \(\sqrt{(5-0)^2 + (0-0)^2} = 5\)
The height can be calculated as the straight line distance between the points (2,3) & (2,0) which lies on the x-axis.
The distance between the points(height) \(\sqrt{(2-2)^2 + (3-0)^2} = 3\)
Therefore, the area of the triangle is \(\frac{1}{2} * 5 * 3 = \frac{15}{2} = 7.5\)
Hope this helps you!
