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# Find area: Given 3 coordinates of a triangles

Registered User Posts: 169 Junior Member
edited August 2009
In the xy-coordinate plane, three vertices of a triangle are (1,2), (7,5), and (7,9). What is the area of the figure?

This is a grid in question. I had to use the distance formula to find the distance of all three sides of the triangle and then use Heron's formula to find the area.

S=(1/2)(A+B+C)
A=[S(S-A)(S-B)(S-C)]^(1/2)

This problem took me about 5 minutes because my method is so tedious and I easily make errors. Can anyone tell me if there is a shortcut to the problem?
Post edited by ajsz033 on

## Replies to: Find area: Given 3 coordinates of a triangles

• Registered User Posts: 1,026 Senior Member
Was this problem from the College Board? Or is it from like Kaplan, Princeton Review...
• Registered User Posts: 359 Member
In fact, there is.

Took about a minute to do using my method. Basically you should sketch out the triangle with points and coordinates, but do not really follow any scale, just mark coordinates. Then use the following formula: S=1/2 base*height to find area of triangle. Also, you will have to introduce another point (7,2) and first find the area of the triangle with points (1,2),(7,2),(7,9). The area of this large triangle is S=1/2*(7-1)*(9-2)=21. Now find the area of triangle with points (1,2),(7,5),(7,2) using the same method. S=1/2*(7-1)*(5-2)=9. The answer to the original question is the latter subtracted from the former, i.e., 21-9= 12. Is this the correct answer?
• Registered User Posts: 3,447 Senior Member
Here's the simplest, most efficient method

Plot the three coordinates just to get a visual of what you're solving for. The base of the triangle is easy to find since (7, 9) is just 4 vertical units above (7, 5). The height is the distance from (1, 2) to the line x = 7 (imagine a dotted line going from (1, 2) to (7,2) ). Therefore, height = 7 -1 = 6.

Area = 1/2 (base)(height)
= 1/2 (4)(6)
= 12
• Registered User Posts: 359 Member
Lol. You are right. Dont have to do all the extra calculations.
• Registered User Posts: 92 Junior Member
In my opinion , the simplest method to this type of problem would be to use the area of a triangle matrix determinant formula. You can google this method, but you might need to have some background in matrices.
• Registered User Posts: 63 Junior Member
here's the easiest way (in my opinion w/o using the calculations and crap) Convert the triangle into a Rectangle by just drawing a rectangle around the triangle (the vertices of triangle should be ON (not within) the rectangle. You will get 4 triangles (three of them will be right triangles) Find their formula and subtract it from the rectangle's formula.... you have the triangles area. Believe me it sounds crazy and all complicated but try it out. PM Me if you still don't get it!
• Registered User Posts: 359 Member
^ I think its in fact the longest method specified thus far. Even longer the the Heron's formula. Jamesford has specified the quickest and easiest method. You simply cannot get any more straightforward - just one half of base * height.
• Registered User Posts: 582 Member
JamesFord's method is the correct one... I've never seen CB given this kind of problem without it being a right triangle....
• - Posts: 1,520 Senior Member
Whenever you see a triangle problem like this, draw it out: the triangle will never be complex. I don't think the SAT has any math questions on it that require memorizing Hero's formula. You can easily solve this question because 2 of the points lie on the same x-position. Jame's method written above is the best.
• Registered User Posts: 33 Junior Member
Lets call the points A B C

Find the vector going from A to B, we'll call it V
Find the vector going from A to B, we'll call it U

Find the cross product of V and U

Find the length of the cross product and divide by 2 and DONE!

yeah seems like a long process but it can be done really quick on a calculator
• Registered User Posts: 582 Member
You really shouldn't use your calculator on this... just sketch it out and the rest is easy... takes like 10 seconds....
• Registered User Posts: 1,316 Senior Member
This took me about 30 seconds in my head - there's one very simple thing that sticks out in the question that makes it faster to do.

So we know that the area of a triangle is base x height/2. Notice that two of the points on the triangle have a common x-coordinate (7). So we can use these coordinates to form the base. The length of the base is 9 - 5 = 4, since that is the difference between the y-coordinates of the two points.

The next thing is to find the height, which is simply 7 minus the x co-ordinate of the third point - i.e. 7 - 1 = 6.

Now we do 6 * 4/2 = 12.

It might be worth imagining the triangle in your head or doing a rough sketch. Also note that I kind of turned the graph around to form the base and height of the triangle.
This discussion has been closed.