What is the area of the triangle formed by the lines y = 7 , x = 8 and 4x+ 3y =7?

To determine the area of the triangle formed intersecting
the 3 given lines, we'll have to determine the length of the sides of the
triangle.

We'll determine the vertices of the triangle
first. The vertices of the triangles are the intercepting point of the given
lines.

We'll determine the first vertex, namely the
intercepting point of x=8 and 4x + 3y = 7.

We'll solve the
system formed by the
equations

x=8

4x + 3y =
7

We'll substitute x=8 in 4x + 3y =
7.

32 + 3y = 7

We'll subtract
32 both sides:

3y = 7 - 32

3y
= 25

y = 25/3

So the first
vertex is the point (8 , 25/3).

We'll determine the next
vertex for y = 7 and 4x + 3y = 7.

We'll substitute y = 7 in
4x + 3y = 7.

4x + 21 = 7

We'll
subtract 21:

4x = 7-21

4x =
14

x = 14/4

x =
7/2

The next vertex is (7/2 ,
7)

The 3rd vertex is
(8,7).

Since x is perpendicular to y, the triangle is right
angled and the area of a right angled triangle is the half-product of the
cathetus.

The base of triangle, AB, one cathetus, is
located between (8,7) and (7/2 , 7).

The length of the base
is:

AB = sqrt[(8-7.2)^2 +
(7-7)^2]

AB =
0.8

The height of triangle, AB, one
cathetus, is located between (8,7) and (8 , 25/3).

AC =
sqrt [(8-8)^2+(25/3 - 7)^2]

AC =
2/7

The area of triangle ABC
is:

A = AB*AC/2

A =
(8*2)/2*10*7

A = 4/5*7

A =
4/35

A = 114/1000 square
units