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IAL 2026 Jan Q4

A Level / Edexcel / P1

IAL 2026 Jan Paper · Question 4

题目

Problem

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

The point A(1,8)A(-1,8) and the point B(7,2)B(7,2) lie on the line L1L_1.

(a) Find an equation for L1L_1 in the form ax+by+c=0ax+by+c=0, where aa, bb and cc are integers.

(3)

The line L2L_2

  • passes through AA and is perpendicular to L1L_1
  • cuts the xx-axis at the point CC

(b) Find the coordinates of CC.

(3)

(c) Hence find the area of triangle ABCABC.

(3)

解答

(a)

解法一

思路

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先用两点求斜率,再用点斜式写直线方程,最后整理成整数系数的 ax+by+c=0ax+by+c=0

答题过程

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The gradient of L1L_1 is

m=287(1)=68=34\begin{align*} m =&\, \frac{2-8}{7-(-1)}\\[3mm] =&\, \frac{-6}{8}\\[3mm] =&\, -\frac34 \end{align*}

Using point B(7,2)B(7,2),

y2=34(x7)\begin{align*} y-2=-\frac34(x-7) \end{align*}

Multiply by 44:

4y8=3x+213x+4y29=0\begin{align*} 4y-8 =&\, -3x+21\\[3mm] 3x+4y-29 =&\, 0 \end{align*}

Therefore

L1:3x+4y29=0\begin{align*} L_1:\quad 3x+4y-29=0 \end{align*}

(b)

解法一

思路

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垂直直线的斜率互为负倒数。L1L_1 的斜率是 34-\dfrac34,所以 L2L_2 的斜率是 43\dfrac43。再令 y=0y=0xx 轴交点。

答题过程

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Since L2L_2 is perpendicular to L1L_1, its gradient is

43\begin{align*} \frac43 \end{align*}

Using A(1,8)A(-1,8),

y8=43(x+1)\begin{align*} y-8=\frac43(x+1) \end{align*}

At the xx-axis, y=0y=0, so

08=43(x+1)24=4(x+1)6=x+1x=7\begin{align*} 0-8 =&\, \frac43(x+1)\\[3mm] -24 =&\, 4(x+1)\\[3mm] -6 =&\, x+1\\[3mm] x =&\, -7 \end{align*}

Therefore

C=(7,0)\begin{align*} C=(-7,0) \end{align*}

(c)

解法一

思路

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因为 ABABL1L_1 上,ACAC 在垂直的 L2L_2 上,所以三角形 ABCABCAA 处是直角。面积是 12ABAC\dfrac12 AB\cdot AC

答题过程

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Find ABAB:

AB=(7(1))2+(28)2=82+(6)2=10\begin{align*} AB =&\, \sqrt{(7-(-1))^2+(2-8)^2}\\[3mm] =&\, \sqrt{8^2+(-6)^2}\\[3mm] =&\, 10 \end{align*}

Find ACAC:

AC=(7(1))2+(08)2=(6)2+(8)2=10\begin{align*} AC =&\, \sqrt{(-7-(-1))^2+(0-8)^2}\\[3mm] =&\, \sqrt{(-6)^2+(-8)^2}\\[3mm] =&\, 10 \end{align*}

Since ABAB is perpendicular to ACAC,

Area=12(AB)(AC)=12(10)(10)=50\begin{align*} \text{Area} =&\, \frac12(AB)(AC)\\[3mm] =&\, \frac12(10)(10)\\[3mm] =&\, 50 \end{align*}

解法二

思路

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鞋带定理(Shoelace Formula)。 对于平面直角坐标系中已知三点坐标的三角形,求面积时有一种极为通用且高效的方法——鞋带定理(也称高斯面积公式)。 三角形 ABCABC 的三个顶点为 A(xA,yA)A(x_A, y_A), B(xB,yB)B(x_B, y_B), C(xC,yC)C(x_C, y_C),其面积公式为:

Area=12(xAyB+xByC+xCyA)(yAxB+yBxC+yCxA)\begin{align*} \text{Area} = \frac{1}{2} \left| (x_A y_B + x_B y_C + x_C y_A) - (y_A x_B + y_B x_C + y_C x_A) \right| \end{align*}

在本题中,已知三点坐标为 A(1,8)A(-1, 8), B(7,2)B(7, 2), C(7,0)C(-7, 0)。我们将坐标代入上述公式即可直接计算出三角形面积。这种方法完全不需要计算任何边长,也不依赖于三角形是否为直角三角形,是非常强力的坐标几何求解技巧。

答题过程

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We can use the Shoelace Formula to calculate the area of triangle ABCABC directly from the coordinates of its vertices: A(1,8)A(-1,8), B(7,2)B(7,2), and C(7,0)C(-7,0).

Area=12(xAyB+xByC+xCyA)(yAxB+yBxC+yCxA)=12((1)(2)+7(0)+(7)(8))(8(7)+2(7)+0(1))=12(2+056)(5614+0)=125842=12100=50.\begin{align*} \text{Area} =&\,\, \frac{1}{2} \left| (x_A y_B + x_B y_C + x_C y_A) - (y_A x_B + y_B x_C + y_C x_A) \right|\\[3mm] =&\,\, \frac{1}{2} \left| ((-1)(2) + 7(0) + (-7)(8)) - (8(7) + 2(-7) + 0(-1)) \right|\\[3mm] =&\,\, \frac{1}{2} \left| (-2 + 0 - 56) - (56 - 14 + 0) \right|\\[3mm] =&\,\, \frac{1}{2} \left| -58 - 42 \right|\\[3mm] =&\,\, \frac{1}{2} \left| -100 \right|\\[3mm] =&\,\, 50. \end{align*}