SOLUTION: Find the area of the region bounded by the x and y axes and the line 2x-3y +7=0.

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!
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2x-3y +7=0
Find the x and y intercepts:
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Let y=0:
2x-3(0)+7=0
2x+7=0
2x=-7
x=-7/2
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Let x=0:
2(0)-3y+7=0
-3y=-7
y=7/3
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The region is a right triangle with Base=7/2 and height=7/3
.

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Area=(1/2)(b)(h)
Area=(1/2)(7/2)(7/3)
Area=49/12
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ANSWER: The area of the region is 49/12 square units.
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Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

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

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