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Area of triangle formed by intersections of 3 lines

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Area of triangle formed by intersections of 3 lines

by gmattesttaker2 » Sat May 03, 2014 10:52 am
Hello,

Can you please assist with this:

What is the area of a triangle created by the intersections of the lines x=4, y=5, and y=−3/4x+20?

OA: 96

x = 4 and y = 5 will intersect at the point (4,5). I am thinking that y = -3/4x + 20 means that the slope m = -3/4 and the y-intercept = 20. I am not sure though how to find out where it will intersect the lines x = 4 and y = 5. Can you please explain it? Thanks a lot.

Regards,
Sri
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by Brent@GMATPrepNow » Sat May 03, 2014 11:44 am
gmattesttaker2 wrote:Hello,

Can you please assist with this:

What is the area of a triangle created by the intersections of the lines x = 4, y = 5, and y= −3/4x + 20?


Let's first sketch the lines x = 4 and y = 5
Image

To find the point where y = (-3/4)x + 20 intersects the line x = 4, replace x with 4 to get: y = (-3/4)4 + 20 = 17
So the point of intersection is (4, 17)

To find the point where y = (-3/4)x + 20 intersects the line y = 5, replace y with 5 to get: 5 = (-3/4)x + 20
When we solve for x, we get x = 20
So the point of intersection is (20, 5)

Add this information to our sketch:
Image

From here, we can determine the length of the right triangle's base and height:
Image

Area = (1/2)(base)(height)
= (1/2)(16)(12)
= 96

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by gmattesttaker2 » Sat May 03, 2014 11:46 am
Brent@GMATPrepNow wrote:
gmattesttaker2 wrote:Hello,

Can you please assist with this:

What is the area of a triangle created by the intersections of the lines x = 4, y = 5, and y= −3/4x + 20?
Hello Brent,

Thank you very much for your prompt and excellent explanation.

Best Regards,
Sri



Let's first sketch the lines x = 4 and y = 5
Image

To find the point where y = (-3/4)x + 20 intersects the line x = 4, replace x with 4 to get: y = (-3/4)4 + 20 = 17
So the point of intersection is (4, 17)

To find the point where y = (-3/4)x + 20 intersects the line y = 5, replace y with 5 to get: 5 = (-3/4)x + 20
When we solve for x, we get x = 20
So the point of intersection is (20, 5)

Add this information to our sketch:
Image

From here, we can determine the length of the right triangle's base and height:
Image

Area = (1/2)(base)(height)
= (1/2)(16)(12)
= 96

Cheers,
Brent
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Posts: 641
Joined: Tue Feb 14, 2012 3:52 pm
Thanked: 11 times
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by gmattesttaker2 » Sat May 03, 2014 11:58 am
gmattesttaker2 wrote:
Brent@GMATPrepNow wrote:
gmattesttaker2 wrote:Hello,

Can you please assist with this:

What is the area of a triangle created by the intersections of the lines x = 4, y = 5, and y= −3/4x + 20?



Let's first sketch the lines x = 4 and y = 5
Image

To find the point where y = (-3/4)x + 20 intersects the line x = 4, replace x with 4 to get: y = (-3/4)4 + 20 = 17
So the point of intersection is (4, 17)

To find the point where y = (-3/4)x + 20 intersects the line y = 5, replace y with 5 to get: 5 = (-3/4)x + 20
When we solve for x, we get x = 20
So the point of intersection is (20, 5)

Add this information to our sketch:
Image

From here, we can determine the length of the right triangle's base and height:
Image

Area = (1/2)(base)(height)
= (1/2)(16)(12)
= 96

Cheers,
Brent

Hello Brent,

Thank you very much for your prompt and excellent explanation.

Best Regards,
Sri
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