Question

Find the area of the triangle whose vertices are given below: A(0,0) B(-5,3) C(3,2) The area of triangle ABC is in ------ square units.

          Find the area of the triangle whose vertices are given
below:
A(0,0)    B(-5,3)    C(3,2)
The area of triangle ABC is in ------ square units.

Added by Rachel L.

Calculus: Early Transcendentals

James Stewart 8th Edition

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Solved by Expert Sri K

08/24/2023

Step 1

Find the length of AB, BC, and AC using the distance formula: AB = sqrt((0-(-5))^2 + (0-3)^2) = sqrt(34) BC = sqrt((3-(-5))^2 + (2-3)^2) = sqrt(65) AC = sqrt((3-0)^2 + (2-0)^2) = sqrt(13) Show more…

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Find the area of the triangle whose vertices are given below: A(0,0) B(-5,3) C(3,2) The area of triangle ABC is in ------ square units.

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Transcript

00:01 Given vertices of a triangle are a 0 0, b -5 ,3 and c is 3 ,2.

00:10 These are the given vertices of a triangle.

00:13 Suppose this is our triangle and its points are a ,b ,c.

00:17 Then a coordinate is 0 ,0.

00:20 B is -5 ,3.

00:23 C is 3 ,2.

00:24 If we draw it on the axis, then it will be a point will be here -5 ,3 that it will be somewhat here and c point is 3 ,2 that means it will be here.

00:39 This triangle will be like this on our axis.

00:43 So these are the points and we have to find here the area of a triangle.

00:48 So calculate the length of the sides a ,b ,a ,c and b ,c using the distance formula.

00:53 It is equal to x2 -x1 square plus y2 -y1 square.

01:02 This is the distance formula we will use to calculate the length of the sides.

01:06 First side is a ,b so it's equal to x2 -x1.

01:12 So this is x2 -5 and x1 is 0.

01:15 So minus 5, sorry, minus 5 minus 0 square plus 3 minus 0 its square.

01:28 This will be under root of 25 plus 9 and it is nothing but under root of 34.

01:34 Similarly we will calculate ac which will be under root of 13 and bc which is equal to the under root of 65.

01:42 Then now we will calculate the semi perimeter.

01:48 Semi perimeter is denoted by s...

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