In the xy-coordinate system, like k has y-intercept 12 and an x-intercept greater than zero. If the area of the triangular region enclosed by line k and the two axes is 30, what is the slope of the line k?
A. -12/5
B. -6/5
C. 6/5
D. 3/2
E. 12/5
The OA is A.
Area of triangle = 1/2*b*h
From the prompt, we know 30 = 1/2*b*12 --> Solve: b = 5.
Coordinates of b = (5, 0) and coordinates of h = (0, 12) --> use the slope formula [12-0]/[0-5] = -12/5.
Has anyone another strategic approach to solve this PS question? Regards!
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In the xy-coorcinate system, like k has y-intercept 12 and
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Sionainn@PrincetonReview
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If you are short on time you can pretty quickly narrow it down to A and B by just ballparking. A quick sketch shows the line is going downhill from left to right so you know you are working with a negative slope.
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Sionainn
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Jeff@TargetTestPrep
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We can let the x-intercept be b. Thus the area of the triangle, in terms of b, is ½(b)(12) = 6b. Since the area of the triangle is given to be 30, we have 6b = 30 and thus b = 5. So line k has y-intercept (0, 12) and x-intercept (5, 0). Thus the slope of line k is (0 - 12)/(5 - 0) = -12/5.AAPL wrote:In the xy-coordinate system, like k has y-intercept 12 and an x-intercept greater than zero. If the area of the triangular region enclosed by line k and the two axes is 30, what is the slope of the line k?
A. -12/5
B. -6/5
C. 6/5
D. 3/2
E. 12/5
Answer: A
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