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09 Oct 2022, 09:46
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On the coordinate plane there are 3 points 𝐴(−1, 1), 𝐵(1, −3), and 𝐶(3, 4). What is the area of the triangle connecting the 3 points?
(A) 10
(B) 11
(C) 12
(D) 14
(E) 13
(A) 10
(B) 11
(C) 12
(D) 14
(E) 13
gmatprepper94111
Joined: 17 Jul 2022
Last visit: 23 Nov 2022
Posts: 24
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Location: United States (CA)
Concentration: Finance, Strategy
Schools: Stanford '25
GPA: 3.9
WE:Corporate Finance (Finance: Venture Capital)
13 Oct 2022, 01:14
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=================================
On the coordinate plane there are 3 points 𝐴(−1, 1), 𝐵(1, −3), and 𝐶(3, 4). What is the area of the triangle connecting the 3 points?
(A) 10
(B) 11
(C) 12
(D) 14
(E) 13
=================================
We want to take the area. Area is 0.5 * base * height -- where height is the altitude of the triangle.
I want AB to be the base of my triangle. This means that we want the altitude dropped from C to be our height.
To simplify things, I'm going to move the triangle such that B sits on the origin; this moves the triangle -1 in the x-direction and +3 in the y direction.
A -> (-2, 4)
B -> (0, 0)
C -> (2, 7)
Let's check if we're lucky and it's a right triangle.
-> Slope of AB is -4 / 2 = -2
-> Slope of BC is 7 / 2 = 3.5
-> Nope!
So we have to tackle the altitude in two steps:
(1) Figure out the point D where the continuation of AB intersects a perpendicular line that runs through point C
(2) Take the distance between C and D
Because it runs through the origin, we can just write the equation of AB as y = -2x --> we know this describes our point D
We then can write the equation of CD.
--> The slope of CD is the negative reciprocal of the slope of AB
--> CD runs through C
So we get y - 7 = 0.5(x -2); we can plug in our equation from AB.
\(y-7 = 0.5(x-2)\)
\((-2x)-7 = 0.5(x-2)\)
\(-2x-7=0.5x-1\)
\(-6 = 2.5x\)
\(x = -6/2.5 = -12/5 = -2.4\)
We plug in again for y and get that y = 4.8.
Now we need the distance between C and D
--> C lies at (2, 7)
--> D lies at (-2.4, 4.8)
\(\sqrt{(2 - (-2.4))^2 + (7 - 4.8)^2}\)
\(=\sqrt{(4.4)^2 + (2.2)^2}\)
\(=\sqrt{19.36 + 4.84}\)
\(= \frac{1}{10} * \sqrt{1936 + 484}\)
\(= \frac{1}{10} * \sqrt{2420}\)
\(= \frac{{11 * 2} }{ 10} * \sqrt{5}\)
And we need the distance from A to B:
\(\sqrt{2^2 + 4^2}\)
\(=\sqrt{4 + 16}\)
\(=\sqrt{20}\)
\(= 2\sqrt{5}\)
Then we multiply the distance from A to B by the distance from C to D and then multiply that by 0.5
Area =
\(\frac{{11 * 2} }{ 10} * \sqrt{5} * 2\sqrt{5} * \frac{1}{2}\)
Area = 11
On the coordinate plane there are 3 points 𝐴(−1, 1), 𝐵(1, −3), and 𝐶(3, 4). What is the area of the triangle connecting the 3 points?
(A) 10
(B) 11
(C) 12
(D) 14
(E) 13
=================================
We want to take the area. Area is 0.5 * base * height -- where height is the altitude of the triangle.
I want AB to be the base of my triangle. This means that we want the altitude dropped from C to be our height.
To simplify things, I'm going to move the triangle such that B sits on the origin; this moves the triangle -1 in the x-direction and +3 in the y direction.
A -> (-2, 4)
B -> (0, 0)
C -> (2, 7)
Let's check if we're lucky and it's a right triangle.
-> Slope of AB is -4 / 2 = -2
-> Slope of BC is 7 / 2 = 3.5
-> Nope!
So we have to tackle the altitude in two steps:
(1) Figure out the point D where the continuation of AB intersects a perpendicular line that runs through point C
(2) Take the distance between C and D
Because it runs through the origin, we can just write the equation of AB as y = -2x --> we know this describes our point D
We then can write the equation of CD.
--> The slope of CD is the negative reciprocal of the slope of AB
--> CD runs through C
So we get y - 7 = 0.5(x -2); we can plug in our equation from AB.
\(y-7 = 0.5(x-2)\)
\((-2x)-7 = 0.5(x-2)\)
\(-2x-7=0.5x-1\)
\(-6 = 2.5x\)
\(x = -6/2.5 = -12/5 = -2.4\)
We plug in again for y and get that y = 4.8.
Now we need the distance between C and D
--> C lies at (2, 7)
--> D lies at (-2.4, 4.8)
\(\sqrt{(2 - (-2.4))^2 + (7 - 4.8)^2}\)
\(=\sqrt{(4.4)^2 + (2.2)^2}\)
\(=\sqrt{19.36 + 4.84}\)
\(= \frac{1}{10} * \sqrt{1936 + 484}\)
\(= \frac{1}{10} * \sqrt{2420}\)
\(= \frac{{11 * 2} }{ 10} * \sqrt{5}\)
And we need the distance from A to B:
\(\sqrt{2^2 + 4^2}\)
\(=\sqrt{4 + 16}\)
\(=\sqrt{20}\)
\(= 2\sqrt{5}\)
Then we multiply the distance from A to B by the distance from C to D and then multiply that by 0.5
Area =
\(\frac{{11 * 2} }{ 10} * \sqrt{5} * 2\sqrt{5} * \frac{1}{2}\)
Area = 11
22 Oct 2022, 23:28
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On the examination, use the plastic notebook to write a huge coordinate system and pin point the points mentioned above.
Then, once you find the triangle, find out the area of the rectangle. What is mean is the complete length of the Y axis that the triangle takes up and the length of the X axis.
Now, the area comes up to be 28. 7 x 4.
When our triangle is enclosed within the rectangle, there are 3 other triangles formed. Each of them have easily measurable lenghts. Because they are parallel to the X and Y axis. Find the area of the triangles. 28 - 4 - 7 - 6 = 11.
Took me a little less than 3 minutes but as long as we figure out a solid way to find the correct answer... why not?
Then, once you find the triangle, find out the area of the rectangle. What is mean is the complete length of the Y axis that the triangle takes up and the length of the X axis.
Now, the area comes up to be 28. 7 x 4.
When our triangle is enclosed within the rectangle, there are 3 other triangles formed. Each of them have easily measurable lenghts. Because they are parallel to the X and Y axis. Find the area of the triangles. 28 - 4 - 7 - 6 = 11.
Took me a little less than 3 minutes but as long as we figure out a solid way to find the correct answer... why not?
