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IAL 2024 Jan Q5

A Level / Edexcel / P1

IAL 2024 Jan Paper · Question 5

题目

Problem

Figure 1 shows a straight line l1l_1 that passes through the points P(−2,9)P(-2,9) and Q(10,6)Q(10,6).

Figure 1

(a) Find an equation of l1l_1 in the form y=mx+cy=mx+c, where mm and cc are constants to be found.

(3)

The straight line l2l_2 passes through the origin OO and is perpendicular to l1l_1.

The lines l1l_1 and l2l_2 meet at the point RR.

(b) Find the coordinates of RR.

(3)

(c) Find the exact area of triangle OPQOPQ.

(3)

解答

(a)

解法一

思路

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先用两点求斜率，再把其中一个点代入 y=mx+cy=mx+c 求截距。

答题过程

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The gradient of l1l_1 is

m= 6−910−(−2)= −312= −14.\begin{align*} m =&\,\frac{6-9}{10-(-2)}\\ =&\,\frac{-3}{12}\\ =&\,-\frac14. \end{align*}

So

y=−14x+c.\begin{align*} y=-\frac14x+c. \end{align*}

Using P(−2,9)P(-2,9),

9= −14(−2)+c9= 12+cc= 172.\begin{align*} 9=&\,-\frac14(-2)+c\\ 9=&\,\frac12+c\\ c=&\,\frac{17}{2}. \end{align*}

Therefore

y=−14x+172.\begin{align*} y=-\frac14x+\frac{17}{2}. \end{align*}

(b)

解法一

思路

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垂直直线的斜率相乘为 −1-1。l1l_1 的斜率是 −14-\frac14，所以 l2l_2 的斜率是 44。又因为 l2l_2 过原点，所以方程是 y=4xy=4x。交点 RR 由两条直线联立求出。

答题过程

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Since l2l_2 is perpendicular to l1l_1, its gradient is 44.

As l2l_2 passes through the origin,

l2:y=4x.\begin{align*} l_2:\quad y=4x. \end{align*}

At RR, the two equations are equal:

4x= −14x+17216x= −x+3417x= 34x= 2.\begin{align*} 4x=&\,-\frac14x+\frac{17}{2}\\ 16x=&\,-x+34\\ 17x=&\,34\\ x=&\,2. \end{align*}

Then

y=4(2)=8.\begin{align*} y=4(2)=8. \end{align*}

Therefore

R=(2,8).\begin{align*} R=(2,8). \end{align*}

(c)

解法一

思路

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点 RR 在 PQPQ 上，也在从原点作出的垂线 OROR 上。所以 OROR 是三角形 OPQOPQ 以 PQPQ 为底的高。面积可以用 12×PQ×OR\frac12\times PQ\times OR。

答题过程

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Since RR lies on PQPQ and OROR is perpendicular to PQPQ, the area of triangle OPQOPQ is

12×PQ×OR.\begin{align*} \frac12\times PQ\times OR. \end{align*}

Now

PQ= (10−(−2))2+(6−9)2= 122+(−3)2= 153= 317,\begin{align*} PQ =&\,\sqrt{(10-(-2))^2+(6-9)^2}\\ =&\,\sqrt{12^2+(-3)^2}\\ =&\,\sqrt{153}\\ =&\,3\sqrt{17}, \end{align*}

and

OR= 22+82= 68= 217.\begin{align*} OR =&\,\sqrt{2^2+8^2}\\ =&\,\sqrt{68}\\ =&\,2\sqrt{17}. \end{align*}

Therefore

Area= 12(317)(217)= 51.\begin{align*} \text{Area} =&\,\frac12(3\sqrt{17})(2\sqrt{17})\\ =&\,51. \end{align*}

解法二

思路

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也可以直接用坐标面积公式。这个方法不需要先找 RR，但要小心代入顺序和绝对值。

答题过程

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Using the coordinate area formula for O(0,0)O(0,0), P(−2,9)P(-2,9) and Q(10,6)Q(10,6),

Area= 12∣0(9−6)+(−2)(6−0)+10(0−9)∣= 12∣−12−90∣= 51.\begin{align*} \text{Area} =&\,\frac12 \left| 0(9-6)+(-2)(6-0)+10(0-9) \right|\\ =&\,\frac12|-12-90|\\ =&\,51. \end{align*}