Question 1012913: Find the area of the region bounded by the x and y axes and the line 2x-3y +7=0. Found 2 solutions by macston, solver91311:Answer by macston(5194) (Show Source):
You can put this solution on YOUR website! .
2x-3y +7=0
Find the x and y intercepts:
.
Let y=0:
2x-3(0)+7=0
2x+7=0
2x=-7
x=-7/2
.
Let x=0:
2(0)-3y+7=0
-3y=-7
y=7/3
.
The region is a right triangle with Base=7/2 and height=7/3
.
.
Area=(1/2)(b)(h)
Area=(1/2)(7/2)(7/3)
Area=49/12
.
ANSWER: The area of the region is 49/12 square units.
.
The given line and the two axes form a right triangle with the right angle at the origin. The absolute value of the -coordinate of the -intercept is equal to the measure of one of the legs of the triangle. The absolute value of the -coordinate of the -intercept is equal to the measure of the other leg of the triangle. The product of the measures of the two legs of a right triangle divided by 2 is the area of the triangle.
Step 1: Substitute 0 for in the given equation and solve for . This is the -coordinate of the -intercept. Take the absolute value of this value.
Step 2: Substitute 0 for in the given equation and solve for . This is the -coordinate of the -intercept. Take the absolute value of this value.
Step 3: Multiply the result of step 1 times the result of step 2. Divide this product by 2.
John
My calculator said it, I believe it, that settles it
