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

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==Solution==
 
==Solution==
  
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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:
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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>.
 +
 
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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>
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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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 +
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{{AHSME 50p box|year=1958|num-b=6|num-a=8}}
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{{MAA Notice}}

Latest revision as of 05:09, 3 October 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:

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

\[-1=2x+2\]

\[2x = -3\]

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

See Also

1958 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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