What is the area of the triangle formed by the intersection of lines y

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

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.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2017/0 ... ties-gmat/

Taking two equations at a time and solving them by setting them equal to each other (since they all return the value of y):

Eq. 1 = Eq. 2
\(2x - 2 = \frac{-x}{2} +8\\
x = 4\)

Substituting the value of x in Eq. 1
\(2x - 2 = y\\
2(4) - 2 = 6 = y\)

Same way,

Eq. 1 = Eq. 3
\(2x - 2 = 0\\
x = 1\)

Eq. 2 and Eq.3
\(\frac{-x}{2} + 8 = 0\\
x = 16\)

Note, we don't need to substitute the value of y back into either of the equations, since y = 0 is given. Thus, the other two points of intersection will be (1,0) and (16,0).

Now, using the formulas below we can find the area of the triangle:

\(Area = \frac{1}{2} ∣x_1(y_2−y_3)+x_2(y_3−y_1)+x_3(y_1−y_2)∣\)

\(Area = \frac{1}{2} ∣4(0−0)+1(0−6)+16(6−0)∣\)

Calculating each term:

4 (0-0) = 0
1(0-6) = -6
16(6-0) = 16*6 = 96

Thus, Area = 45, op. D is the answer.