Find the area of the triangle formed by the intersection of the lines:?

Question

A. x + y = 8

B. y = 1/2x -1

C. x = -2

Answer 1

draw all lines first and separate into 5 sub-area

(i) ----> 8 - def-int(-2 to +2)(x/2 - 1) = 8 - (4) = 4

(ii) ---> 6*2 = 12

(iii) --> 4 - def-int(8 to 10)(-x + 8) = 4 - 2 = 2

(iv) --> def-int(2 to 6)(x/2 - 1) = 3 - 1 = 2

(v) --> def-int(6 to 8)(-x+8) = 4 - 2 = 2

Total area = (i) + (ii) + (iii) + (iv) + (v) = 4 + 12 + 2 + 2 + 2 = 22

Answer 2

Let
x = -2 be the base of the triangle.

Now find where it intersects the other two lines to calculate its length.

x + y = 8
-x + y = 8
y = 10

y = (1/2)x - 1 = (1/2)(-2) - 1 = -1 - 1 = -2

So the length b, of the base is:

b = 10 - (-2) = 12

Now find the height h, of the triangle. To do that we need to find the vertex of the triangle at the intersection of the first two lines.

x + y = 8
y = 1/2x - 1

Use substitution.

x + (1/2)x - 1 = 8
(3/2)x = 9
x = 6

The height h, of the triangle is:

h = 6 - (-2) = 8

So the area of the triangle is:

Area = (1/2)bh = (1/2)*12*8 = 48