Triangle ABC On the Coordinate Plane

Set 1.4
Question 34

See attachment.

Work out (a) through (c).

 

a.
the area of triangle is equal to area of rectangle minus the area of three right triangles


A=(3,4)
B=(8,5)
C=(7,8}

sides of rectangle are:
from A to vertical side contains B, distance is 5 units (the length of rectangle)
from A to vertical side contains C, distance is 4 units (the width of rectangle)

area of rectangle is 5*4=20

right triangle with hypotenuse AC has an area of (4*4)/2=16/2=8
right triangle with hypotenuse AB has an area of (5*1))/2=5/2=2.5
right triangle with hypotenuse BC has an area of (3*1))/2=3/2=1.5

add all: 8+2.5+1.5=12

then area of triangle ABC is: 20-12=8

b.

A=(1,3)
B=(4,1)
C=(10,4)

horizontal distance from A to C is 9 units
vertical distance from B to C is 3 units
sides of rectangle are:9 and 4
area of rectangle is: 9*3=27


right triangle with hypotenuse AC has an area of (9*1)/2=9/2
right triangle with hypotenuse AB has an area of (3*2))/2=3
right triangle with hypotenuse BC has an area of (6*3))/2=9

then area of triangle ABC is: 27-(9/2+3+9)=10.5

c)
using formula (1/2)(x1*y2-x2*y1 +x2*y3-x3*y2+x3*y1-x1*y3)

A=(1,3)=(x1,y1)
B=(4,1)=(x2,y2)
C=(10,4)=(x3,y3)

area of triangle ABC is:
(1/2)(x1*y2-x2*y1 +x2*y3-x3*y2+x3*y1-x1*y3)
=(1/2)(1*1-4*3 +4*4-10*1+10*3-1*4)

=(1/2)(21)

=10.5

 

I will post questions like this one from time to time. Due to lack of time, I don't have time to completely deep into a problem like this one but at least I have a legitimate reason: working full-time.