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We are supposed to find the area of the figure enclosed by the lines x=1, y=1 and y=6-x which is nothing but Area of Triangle ABC. AB = 4 (distance between (1,1) and (5,1)) AC = 4 (distance between (1,1) and (1,5))

Area of Triangle ABC which is right angled at A = (1/2) * AB * AC = (1/2)*4*4 = 8 So, Answer A

Hope it helps!

PawanRamnani1234 wrote:

What is the area inscribed by the lines y =1, x = 1, y = 6-x on an xy-coordinate plane?

a) 8 b) 10 c) 12 d) 14 e) 18

Looking for appropriate solution for above problem.

What is the area inscribed by the lines y =1, x = 1, y = 6-x on an xy-coordinate plane?

a) 8 b) 10 c) 12 d) 14 e) 18

First, let's graph the lines y = 1 and x = 1

At this point, we need to find the points where the line y = 6-x INTERSECTS the other two lines.

For the vertical line, we know that x = 1, so we'll PLUG x = 1 into the equation y = 6-x to get y = 6-1 = 5 Perfect, when x = 1, y = 5, so one point of intersection is (1,5)

For the horizontal line, we know that y = 1, so we'll PLUG y = 1 into the equation y = 6-x to get 1 = 6-x. Solve to get: x = 5 So, when y = 1, x = 5, so one point of intersection is (5,1)

Now add these points to our graph and sketch the line y = 5-x

At this point, we can see that we have the following triangle.

The base has length 4 and the height is 4 Area = (1/2)(base)(height) = (1/2)(4)(4) = 8

What is the area inscribed by the lines y =1, x = 1, y = 6-x on an xy-coordinate plane?

a) 8 b) 10 c) 12 d) 14 e) 18

Hi,

The sketches have been made above, but few points.. 1) Three equation,so three lines and therefore a triangle..two sides x=1and y=1 make a right angle so we are talking of a RIGHT angle triangle.. 2) the third line will intersect these two lines .. On line x=1, when x is 1 in y=6-x, y=6-1=5.. similarly on line y=1at x=5.. 3) therefore the length of each side is 5-1=4.. Area =1/2*4*4=8.. A
_________________

Re: What is the area inscribed by the lines y =1, x = 1, y = 6-x on an xy- [#permalink]

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16 Jun 2017, 01:02

PawanRamnani1234 wrote:

What is the area inscribed by the lines y =1, x = 1, y = 6-x on an xy-coordinate plane?

a) 8 b) 10 c) 12 d) 14 e) 18

There's no need to actually diagram this problem- simply not much time on the GMAT; however, it might be ok in the beginning. Anyways, the concept behind this question is simply testing your ability to plug in the given y and x values in order to find the two other vertices of the figure- which if you do diagram you'll see is an isosceles equilateral triangle.

1= 6-x -5=-x 5=x (we're not done yet though)

y=6-(x) y=6- 1 y= 5 ( we are done yet either)

x and y intersect at (1,1) so actually each length of this triangle would be 4= 4 spots upwards from (1,1) actually gives us (1,5) - the vertice of the triangle derived from plugging in our x value; And 4 spots to the right of (1,1) gives us (5,1) the vertice of the triangle derived from plugging in the y value. And because this triangle is a right triangle (the intersection of two perfectly straight lines, lines with a slope of 0, is 90 degrees and the lines x and y are also perpendicular) the area can be calculated simply by