题目

Problem

The line $l_1$ has equation $3y-2x=30$ .

The line $l_2$ passes through the point $A(24,0)$ and is perpendicular to $l_1$ .

Lines $l_1$ and $l_2$ meet at the point $P$ .

(a) Find, using algebra and showing your working, the coordinates of $P$ .

(5)

Given that $l_1$ meets the $x$ -axis at the point $B$ ,

(b) find the area of triangle $BPA$ .

(3)

解答

展开

先求 $l_1$ 的斜率，再用垂直关系得到 $l_2$ 的斜率。写出 $l_2$ 后，与 $l_1$ 联立求交点。

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For $l_1$ ,

\begin{align*} 3y-2x=&\,30\\ 3y=&\,2x+30\\ y=&\,\frac23x+10. \end{align*}

So the gradient of $l_1$ is $\frac23$ . Since $l_2$ is perpendicular to $l_1$ , its gradient is $-\frac32$ .

Using $A(24,0)$ ,

\begin{align*} y-0=&\,-\frac32(x-24)\\ y=&\,-\frac32x+36. \end{align*}

At $P$ , the two line equations are equal:

\begin{align*} \frac23x+10=&\,-\frac32x+36\\ 4x+60=&\,-9x+216\\ 13x=&\,156\\ x=&\,12. \end{align*}

Then

\begin{align*} y=&\,\frac23(12)+10\\ =&\,18. \end{align*}

Therefore

\begin{align*} P=(12,18). \end{align*}

展开

先求 $B$ 。因为 $B$ 在 $x$ 轴上，所以 $y=0$ 。然后用 $BA$ 作底，高就是 $P$ 到 $x$ 轴的距离。

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At $B$ , $y=0$ on $l_1$ :

\begin{align*} 3(0)-2x=&\,30\\ x=&\,-15. \end{align*}

\begin{align*} B=(-15,0). \end{align*}

The base $BA$ has length

\begin{align*} 24-(-15)=39. \end{align*}

The height of the triangle is the $y$ coordinate of $P$ , which is $18$ . Therefore

\begin{align*} \text{Area} =&\,\frac12(39)(18)\\ =&\,351. \end{align*}