题目

Problem

The straight line $l_1$ shown in Figure 2 has equation $5y=2x+10$ .

Figure 2

The points $A$ and $B$ lie on $l_1$ such that

point $A$ lies on the $y$ -axis
point $B$ has $x$ coordinate $10$

(a) Find the distance $AB$ writing your answer as a fully simplified surd.

(3)

The straight line $l_2$ also shown in Figure 2

passes through $B$
is perpendicular to $l_1$

(b) Find an equation for $l_2$ writing your answer in the form $ax+by+c=0$ , where $a$ , $b$ and $c$ are integers.

(4)

Line $l_2$ crosses the $x$ -axis at the point $C$ .

Point $D$ is such that the points $A$ , $B$ , $C$ and $D$ form the vertices of a rectangle, shown shaded in Figure 2.

(3)

解答

(a)

解法一

思路

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先求 $A$ 和 $B$ 的坐标。 $A$ 在 $y$ 轴上，所以 $x=0$ ； $B$ 的 $x$ 坐标是 $10$ 。再用两点距离公式。

答题过程

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For $A$ , set $x=0$ :

\begin{align*} 5y=&\,2(0)+10\\ y=&\,2. \end{align*}

\begin{align*} A=(0,2). \end{align*}

For $B$ , set $x=10$ :

\begin{align*} 5y=&\,2(10)+10\\ 5y=&\,30\\ y=&\,6. \end{align*}

\begin{align*} B=(10,6). \end{align*}

Therefore

\begin{align*} AB =&\,\sqrt{(10-0)^2+(6-2)^2}\\ =&\,\sqrt{100+16}\\ =&\,\sqrt{116}\\ =&\,2\sqrt{29}. \end{align*}

(b)

解法一

思路

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先把 $l_1$ 写成 $y=\frac25x+2$ ，所以斜率是 $\frac25$ 。垂线斜率是 $-\frac52$ 。再用点 $B(10,6)$ 写直线方程。

答题过程

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From

\begin{align*} 5y=2x+10, \end{align*}

we get

\begin{align*} y=\frac25x+2. \end{align*}

So the gradient of $l_1$ is $\frac25$ , and the gradient of $l_2$ is

\begin{align*} -\frac52. \end{align*}

Using $B(10,6)$ ,

\begin{align*} y-6=&\,-\frac52(x-10). \end{align*}

Multiply by $2$ :

\begin{align*} 2y-12=&\,-5x+50\\ 5x+2y-62=&\,0. \end{align*}

Therefore

\begin{align*} 5x+2y-62=0. \end{align*}

(c)

解法一

思路

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矩形面积是 $AB\times BC$ 。已经有 $AB=2\sqrt{29}$ ，再由 $l_2$ 的 $x$ 截距求 $C$ ，然后用距离公式求 $BC$ 。

答题过程

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At $C$ , $y=0$ on $l_2$ :

\begin{align*} 5x+2(0)-62=&\,0\\ x=&\,\frac{62}{5}. \end{align*}

\begin{align*} C=\left(\frac{62}{5},0\right). \end{align*}

Now

\begin{align*} BC =&\,\sqrt{\left(\frac{62}{5}-10\right)^2+(0-6)^2}\\ =&\,\sqrt{\left(\frac{12}{5}\right)^2+36}\\ =&\,\sqrt{\frac{144}{25}+\frac{900}{25}}\\ =&\,\sqrt{\frac{1044}{25}}\\ =&\,\frac{6\sqrt{29}}{5}. \end{align*}

Hence the area is

\begin{align*} AB\cdot BC =&\,2\sqrt{29}\cdot\frac{6\sqrt{29}}{5}\\ =&\,\frac{12\cdot29}{5}\\ =&\,\frac{348}{5}\\ =&\,69.6. \end{align*}

IAL 2024 May R Q7

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程