SOLUTION: Find the equation of the line with slope -12/5 crosses the first quadrant and forms with the axes a triangle with the perimeter of 15 units. a. 12x

SOLUTION: Find the equation of the line with slope -12/5 crosses the first quadrant and forms with the axes a triangle with the perimeter of 15 units. a. 12x - 5y = 30 b. 12x + 5y = 30

Question 1189324: Find the equation of the line with slope -12/5 crosses the first quadrant and forms with the axes a triangle with
the perimeter of 15 units.
a. 12x - 5y = 30
b. 12x + 5y = 30
c. -12x + 5y = 30
d. 12x + 5y = -30
Found 2 solutions by Alan3354, greenestamps:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Find the equation of the line with slope -12/5 crosses the first quadrant and forms with the axes a triangle with
the perimeter of 15 units.
a. 12x - 5y = 30
b. 12x + 5y = 30
c. -12x + 5y = 30
d. 12x + 5y = -30
----------------------
a & c have positive slopes
d does not cross Q1
=====================
b is the only possibility of the 4.
I didn't check for the perimeter.
Answer by greenestamps(13203) (Show Source): You can put this solution on YOUR website!

As the other tutor points out, the question as posed, with four answer choices, is easily solved because only one of the choices is possible. For the problem to be written in a way that makes the student do some actual problem solving, there would have to be a 5th answer choice "e. none of the above".

So let's go ahead and show that answer choice b actually gives a triangle with perimeter 15.

With the line having a slope of -12/5, the x- and y-intercepts are of the form (5a,0) and (0,12a). So the lengths of the two legs of the right triangle are 5a and 12a; that makes the length of the hypotenuse 13a.

The perimeter is then 5a+12a+13a=30a; since the perimeter is 15, a=1/2, and the x- and y-intercepts are (5/2,0) and (0,6).

Now we have a line with slope -12/5 and y-intercept 6:

y=(-12/5)x+6
(12/5)x+y=6
12x+5y=30

Yes, answer choice b is indeed correct.

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