题目

Problem

The line $l_1$ has equation $3x+5y-7=0$ .

(a) Find the gradient of $l_1$ .

(2)

The line $l_2$ is perpendicular to $l_1$ and passes through the point $(6,-2)$ .

(b) Find the equation of $l_2$ in the form $y=mx+c$ , where $m$ and $c$ are constants.

(3)

解答

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把直线方程整理成 $y=mx+c$ ， $x$ 的系数就是斜率。

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\begin{align*} 3x+5y-7=&\,0\\ 5y=&\,-3x+7\\ y=&\,-\frac35x+\frac75. \end{align*}

So the gradient is

\begin{align*} -\frac35. \end{align*}

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垂直线的斜率互为负倒数，所以 $l_2$ 的斜率是 $\frac53$ 。再代入点 $(6,-2)$ 求截距。

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The gradient of $l_2$ is

\begin{align*} \frac53. \end{align*}

Using $(6,-2)$ ,

\begin{align*} y+2=&\,\frac53(x-6)\\ y+2=&\,\frac53x-10\\ y=&\,\frac53x-12. \end{align*}