Difference between revisions of "1958 AHSME Problems/Problem 7"
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The slope of the line is <math> \frac{ \Delta y }{ \Delta x} = \frac{9-1}{3-(-1)} = 2</math>. Using the formula for the point-slope form of a line, we have <math> y-y_1 = m(x-x_1)</math>, so <math>y-1=2(x-(-1)) \to y-1=2(x+1)</math>. | The slope of the line is <math> \frac{ \Delta y }{ \Delta x} = \frac{9-1}{3-(-1)} = 2</math>. Using the formula for the point-slope form of a line, we have <math> y-y_1 = m(x-x_1)</math>, so <math>y-1=2(x-(-1)) \to y-1=2(x+1)</math>. | ||
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The x-intercept is the x-value when <math>y=0</math>, so we substitute 0 for y: | The x-intercept is the x-value when <math>y=0</math>, so we substitute 0 for y: | ||
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<cmath>x = -\frac{3}{2} \to \boxed{\text{(A)}}</cmath> | <cmath>x = -\frac{3}{2} \to \boxed{\text{(A)}}</cmath> | ||
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+ | ==See Also== | ||
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+ | {{AHSME 50p box|year=1958|num-b=6|num-a=8}} | ||
+ | {{MAA Notice}} |
Latest revision as of 06:09, 3 October 2014
Problem
A straight line joins the points and
. Its
-intercept is:
Solution
The slope of the line is . Using the formula for the point-slope form of a line, we have
, so
.
The x-intercept is the x-value when , so we substitute 0 for y:
See Also
1958 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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All AHSME Problems and Solutions |
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