Answer= D

Step 1: Find the intersection of 2 lines: y = 2x - 2 and y = -x/2 + 8
=> 2x - 2 = -x/2 + 8
=> 4x - 4 = -x + 16
=> 5x = 20
=> x = 4

Put this value in any of the above equations to solve for y
=> y = 2x - 2
=> y = 8 - 2
=> y = 6 ......................... This will be height of triangle

Step 2: Find the base of triangle; Since y = 0 is nothing but X-Axis we need to find the distance between the intersection of 2 lines with x axis
Line 1 -->
=>2x-2 = 0
=> x = 1

Line 2 -->
=> x/2 = 8
=> x = 16

Length of base = 16 - 1 = 15

Area of triangle = 1/2 * 15 * 6 = 15 * 3 = 45

I got \(S=\frac{1}{2}*\frac{4}{7}*\frac{1}{3}=\frac{2}{21}\).

Please correct me if I'm wrong!
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Bunuel please provide the solution for this. i got the coordinates as (1,0), (2/3,0) and (5/7, -4/7).
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Not even close to the answer. Expert guidance neeeded?

I am getting 2/21. Please re-check the answer choices and confirm

Sorry everyone. There was a typo in the question. Now it's edited.
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The lines ( other than the x axis ) intersect at the point (4,6). (Found on solving the two lines two equations)
The required area is formed by the triangle with vertex at (4,6) it has a height of 6.

The two base points are (1,0) & (16,0) .... Found on checking where the lines intersect the x axis ( substitute y=0)

Hence base is of length 15. The area is 15*6*0.5 = 45.

Hence Option (D) is correct.

Regards,
Gladi
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Given three lines with equations as (i) \(y = 2x – 2\), (ii) \(y = −\frac{x}{2} + 8\), and (iii) \(y = 0\)

Lets find the point of intersection of each like with \(y = 0 (x axis)\)

\(y = 2x – 2\) intersects x axis at \(x = 1\)

\(y = −\frac{x}{2} + 8\) intersects x axis at \(x = 16\)

The distance between these two intersection points is the base of the triangle \(= 16 - 1 = 15\)

Now the height of the triangle is the y coordinate of the intersection of the two lines (i) & (ii), Solving them we get,

the height \(= 6\)

Hence area of the triangle \(= 1/2*6*15 = 45\)

Answer D.

Thanks,
GyM
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two are perpendicular lines, can we use this bit of information to find the area in some other way??

VeritasKarishma, I need some guidance from you with this problem. Here was my approach to this problem -

I y=2x–2 (slope = 2)
II y=−x/2+8 (slope = -1/2)
III y=0 (it is the x axis; slope = 0)

First thing that I noticed here is that lines 1 and 2 are perpendicular to each other, so the triangle must be a right angled triangle.

Next I found the intersection points of lines I and II on the x-axis.
For line 1 the intersection point on the x axis was (1,0) and for line 2 the intersection on the x-axis was (16,0).

Now since the lines 1 and 2 are perpendicular to each other, then the line segment formed by the intersecting points on the x axis must be the hypotenuse of the right angled triangle formed by the 3 lines. The length of this line segment is 15 (16 -1) and it fits into the 3x, 4x,5x Pythagorean triplet and I concluded that the other two sides must be 9 and 12 respectively.

So I calculated the area to be 1/2 * 9 * 12 = 54.

Could you please explain where I am going wrong with this approach? Thanks for your help !

Think about it: If a right triangle has hypotenuse 15, will the legs necessarily be 9 and 12?
Can the legs be 5 and 10sqrt(2)?
How about 1 and sqrt(224)?
I am sure you can now think of many many cases in which the hypotenuse will be 15.

Pythagorean tripleys give you solutions with all integer sides. They do not give you all solutions.
Here is a post on properties of pythagorean triples:
https://www.veritasprep.com/blog/2017/0 ... ties-gmat/
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Thanks VeritasKarishma, this makes sense. I had incorrectly assumed the other lengths to be integers (and thus Pythagorean triplets) as well since the area of the triangle in all the options were integers. Thanks for explaining this !