What is the area, in square units, of triangle ABC?

Thank you!

AnonymousConfusedGuy Jun 6, 2018

#1**+1 **

We need to find the distance between

(0,6) and (-4,3) = √[ ( -4)^2 + (6-3)^2 ] = √ [4^2 + 3^2 ] =√[25] = 5

Now, ACG....we could find the lengths of the other two sides and use something known as "Heron's Formula" to calculate the area, but it's messy.....here's an easier way....let the first distance we calculated be the base of the triangle

We need to find the equation of the line between the first two points....

The slope is [ 6 -3] / [ 0 - -4] = 3/4

The equation is

y = (3/4)(x - 0) + 6

y = (3/4)x + 6 ....... now.... we need this form : Ax + BY + C = 0....so ....multiply through by 4.....

4y = 3x + 24

3x - 4y + 24 = 0

Now the distance between (2, -2) and this line will be the altitude of the triangle and it is given by

l Ax + By + C l / √[ A^2 + B^2 ] where (x,y) = (2, -2) ....so we have

l 3(2) - 4(-2) + 24 l / √ [ (3)^2 + (-4)^2 ] =

l 6 + 8 + 24 l / √25 =

38 / 5 units

So....the area is

(1/2) base * altitude =

(1/2) (5) * 38/ 5 =

19 units^2

CPhill Jun 6, 2018

#2**+1 **

As Chris mentioned:

I calculated the distances between the points (side lengths) as

5 7.81 and 8.25

thus p = 10.53

sqrt(10.53(10.53-5)(10.53-7.81)(10.53-8.25)) = 19.0 units^2

(just as Chris found!)

ElectricPavlov Jun 6, 2018