In Exercises 45–46, find the area of the triangle with the given ...

Question

In Exercises 45–46, find the area of the triangle with the given vertices. Round to the nearest square unit.(-2, -3), (-2, 2), (2, 1)

Accepted Answer

Hey, everyone in this problem, we're given the vertices of a triangle and we're asked to determine the area of that triangle. The inter writer answer to the nearest integer, we have four answer choices. Option A three, option B 20 option C 40 option D five. So the three vertices we're given, we're gonna go ahead just draw them out. So we get an idea of what it is that we're actually looking for. So the first vertices is negative one comma three. So that's gonna be, we're here. So OK, in the second quadrant and we have negative one comma four. So it's gonna be just above the first point and then we have the final vertices five comma eight. OK? So that's gonna be over in quadrant. And so if we can make these three points together, we have this triangle. And what we are looking for is the area. All right. So let's go ahead and label these points. OK? So we're gonna say that the first point negative one comma three is A, the second point negative one, comma four is B and the 3rd 0.5 comma eight is C and we can label them on our diagram as well. So let's recall, we're looking for an area we were given the coordinates of the vertices. So we can write the area A as one half multiplied by the absolute value of X one multiplied by Y two minus Y three plus X two multiplied by Y three minus Y what plus X three multiplied by Y one minus Y two. So this looks a little messy but what we're essentially doing, we're taking each X value multiplying it by the difference in the other Y values. OK? For each of those three values, now, we've written this in terms of 12 and three. OK. That's typically how it would appear in your textbook. We've labeled our points A B and C. So let's just rewrite this according to A B and C so that it's really clear. So we have our area which is gonna be the one half, the absolute value of X A multiplied by YB. Mine is Y IC plus XB multiplied by YC minus ye plus XC multiplied by Y A minus YB. All right. So we have all three points. We have all of these X and Y values. All we need to do is substitute in our values. So we get that the area A is gonna be equal to one half multiplied by the absolute value A negative one multiplied by four minus eight. OK. So we have the X value of the first point. And then we're subtracting the Y values of the other two points plus the X value of B negative one multiplied by the Y value of C eight minus the Y value of A three. And finally, we have the X value of C which is five multiplied by the Y value of A three minus the Y value of B four. So we have this area now with all of these numbers and we just need to simplify. So we have that the area is one half the absolute value. OK? We have negative one multiplied by four minus eight. So that's gonna be negative one multiplied by negative four plus negative one multiplied by eight minus three. Well, that's gonna be five plus five multiplied by three minus four, which is gonna be negative one. All right, we're gonna keep simplifying area is one half multiplied by the absolute value of negative one multiplied by negative four. It's gonna give us positive four, negative one multiplied by five, gives us negative five. So we have four minus five and then we have five multiplied by negative one again, giving us negative five. And so we have four minus five minus five inside of our absolute volume. This gives us that the area is one half multiplied by the absolute value or minus five minus five, gives us negative six. The absolute value of negative six is just six. So we have one half multiplied by six, which gives us a final area of three. All right. So we were given the three points of a triangle and we found that the area is going to be three, which corresponds with answer choice. A thanks everyone for watching. I hope this video helped see you in the next one.

In Exercises 45–46, find the area of the triangle with the given vertices. Round to the nearest square unit.(-2, -3), (-2, 2), (2, 1)

Key Concepts

Coordinate Geometry

Area of a Triangle

Determinant Method for Area

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