Difference between revisions of "1958 AHSME Problems/Problem 7"

Revision as of 23:17, 3 January 2014

Problem

A straight line joins the points $(-1,1)$ and $(3,9)$ . Its $x$ -intercept is:

$\textbf{(A)}\ -\frac{3}{2}\qquad \textbf{(B)}\ -\frac{2}{3}\qquad \textbf{(C)}\ \frac{2}{5}\qquad \textbf{(D)}\ 2\qquad \textbf{(E)}\ 3$

Solution

The slope of the line is $\frac{ \Delta y }{ \Delta x} = \frac{9-1}{3-(-1)} = 2$ . Using the formula for the point-slope form of a line, we have $y-y_1 = m(x-x_1)$ , so $y-1=2(x-(-1)) \to y-1=2(x+1)$ . The x-intercept is the x-value when $y=0$ , so we substitute 0 for y:

$0-1=2x+2$

$-1=2x+2$

$2x = -3$

$x = -\frac{3}{2} \to \boxed{\text{(A)}}$

@@ Line 11: / Line 11: @@
 ==Solution==
-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>

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Difference between revisions of "1958 AHSME Problems/Problem 7"

Revision as of 23:17, 3 January 2014

Problem

Solution