1964 AHSME Problems/Problem 15
Problem 15
A line through the point cuts from the second quadrant a triangular region with area . The equation of the line is:
Solution
The right triangle has area and base , so the height satisfies . This means . Becuase the triangle is in the second quadrant, the coordinates are the origin, , and , which means the third coordinate is .
So, we want the equation of a line though and . The slope is , which simplifies to .
The y-intercept is , so the line in slope-intercept form is , or:
All the solutions have a positive x-coefficient and no fractions, so we clear the fractions by multiplying by and move the term to the right to get:
This is equivalent to option
See Also
1964 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.