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 x-intercept is the x-value when <math>y=0</math>, so we substitute 0 for y: | + | 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 x-intercept is the x-value when <math>y=0</math>, so we substitute 0 for y: | ||
<cmath>0-1=2x+2</cmath> | <cmath>0-1=2x+2</cmath> |
Revision as of 23:17, 3 January 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: