1958 AHSME Problems/Problem 7

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:

$0-1=2x+2$

$-1=2x+2$

$2x = -3$

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

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1958 AHSME Problems/Problem 7

Problem

Solution