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Algebra Difficulty: Hard

Line $k$ $k$ is shown in the xy-plane. Line $j$ $j$ (not shown) is perpendicular to line $k$ . What is the slope of line $j$ ?

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Explanation

The correct answer is $\frac{6}{5}$ . It’s given that line $j$ is perpendicular to line $k$ in the xy-plane. This means that the slope of line $j$ is the opposite reciprocal of the slope of line $k$ . For a line that passes through the points $left parenthesis x 1 comma y 1 right parenthesis$ and $left parenthesis x 2 comma y 2 right parenthesis$ in the xy-plane, the slope of the line can be calculated as $\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ . It's shown that line $k$ passes through the points $left parenthesis negative 6 comma 0 right parenthesis$ and $left parenthesis 0 comma negative 5 right parenthesis$ in the xy-plane. Substituting $negative 6$ for $x 1$ , $0$ for $y 1$ , $0$ for $x 2$ , and $negative 5$ for $y 2$ in $\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ yields $\frac{- 5 - 0}{0 - (- 6)}$ , or $- \frac{5}{6}$ . The opposite reciprocal of $- \frac{5}{6}$ is $\frac{6}{5}$ . Therefore, the slope of line $j$ is $\frac{6}{5}$ . Note that 6/5 and 1.2 are examples of ways to enter a correct answer.