SOLUTION: find area of the triangle constructed by the lines5x+7y=35,4x+3y=12and x axis

Question 1012550: find area of the triangle constructed by the lines5x+7y=35,4x+3y=12and x axis
Found 2 solutions by Alan3354, macston:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
find area of the triangle constructed by the lines5x+7y=35,4x+3y=12and x axis
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Find the intersection of the 2 lines.
5x+7y=35
4x+3y=12
The y value of the intersection is 80/13
Since the other 2 vertices are on the x-axis, 80/13 is the height of the triangle.
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Find the intersections of the lines and the x-axis
5x + 7y = 35
@y = 0; x = 7
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4x + 3y = 12
@y = 0; x = 3
The base = 7-3 = 4
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Area = b*h/2 = (80/13)*4/2
Area = 160/13 sq units
Answer by macston(5194) (Show Source): You can put this solution on YOUR website!
.
Find vertices
Intersection of 5x+7y=35 (red on graph) and x-axis (purple on graph):
Set y=0, solve for x
.
5x+7(0)=35
5x=35
x=7
One vertex is (7,0)(red crosses purple)
.
Intersection of 4x+3y=12 (green on graph) and x axis (purple on graph):
.
4x+3(0)=12
4x=12
x=3
Second vertex at (3,0) (green crosses purple)
.
The base is from (3,0)to (7,0)
Length of base is (7-3)=4
b=4
.
5x+7y=35 . Multiply this by 3 (red on graph)
4x+3y=12 . Multiply this by 7 (green on graph)
.
15x+21y=105
28x+21y=84 . Subtract second from first
-13x=21
x=-21/13
.
5x+7y=35
5(-21/13)+7y=35
-105/13+7y=35
7y=455/13+105/13
7y=560/13
y=80/13
Third vertex is (-21/13,80/13) (red crosses green)
The height of the triangle is the y-value of this vertex.
h=80/13
.

square units
.

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