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:
  

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:

\[0-1=2x+2\]

\[-1=2x+2\]

\[2x = -3\]

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