题目

Problem

In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.

Given that

(a) show that the gradient of $l_1$ is $p\sqrt3$ , where $p$ is a rational constant to be found. You must show each step of your working.

(2)

The straight line $l_2$ is perpendicular to $l_1$ and passes through $A$ .

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

(3)

解答

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用两点斜率公式。分母会出现 $\sqrt3$ ，所以要有理化，才能写成 $p\sqrt3$ 的形式。

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The gradient of $l_1$ is

\begin{align*} m =&\,\frac{8-5}{7\sqrt3-(-2\sqrt3)}\\ =&\,\frac{3}{9\sqrt3}\\ =&\,\frac{1}{3\sqrt3}. \end{align*}

Rationalise the denominator:

\begin{align*} m =&\,\frac{1}{3\sqrt3}\cdot\frac{\sqrt3}{\sqrt3}\\ =&\,\frac{\sqrt3}{9}. \end{align*}

\begin{align*} m=\frac19\sqrt3, \end{align*}

and therefore

\begin{align*} p=\frac19. \end{align*}

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$l_2$ 垂直于 $l_1$ ，所以斜率是 $l_1$ 斜率的负倒数。由于 $l_1$ 的斜率是 $\frac{\sqrt3}{9}$ ， $l_2$ 的斜率是 $-\frac{9}{\sqrt3}=-3\sqrt3$ 。再代入点 $A$ 写直线方程。

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The gradient of $l_2$ is

\begin{align*} -\frac{1}{\sqrt3/9} =&\,-\frac9{\sqrt3}\\ =&\,-3\sqrt3. \end{align*}

Since $l_2$ passes through $A(-2\sqrt3,5)$ ,

\begin{align*} y-5=&\,-3\sqrt3(x-(-2\sqrt3))\\ y-5=&\,-3\sqrt3(x+2\sqrt3)\\ y-5=&\,-3\sqrt3x-18\\ y=&\,-3\sqrt3x-13. \end{align*}