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IAL 2020 Oct Q6

A Level / Edexcel / P1

IAL 2020 Oct Paper · Question 6

题目

Problem

The points AA and BB have coordinates (−4,11)(-4,11) and (8,2)(8,2) respectively.

(a) Find the gradient of ABAB.

(2)

The line ll is the perpendicular bisector of ABAB.

(b) Find an equation for ll, giving your answer in the form ax+by+c=0ax+by+c=0, where aa, bb and cc are integers to be found.

(4)

The point CC lies on ll such that the area of triangle ABCABC is 37.537.5 square units.

(c) Find the possible coordinates of CC.

(5)

解答

(a)

解法一

思路

展开

用两点斜率公式：纵坐标差除以横坐标差。

答题过程

展开 gradient of AB= 2−118−(−4)= −912= −34.\begin{align*} \text{gradient of }AB =&\,\frac{2-11}{8-(-4)}\\ =&\,\frac{-9}{12}\\ =&\,-\frac34. \end{align*}

(b)

解法一

思路

展开

垂直平分线经过 ABAB 的中点，并且斜率是 ABAB 斜率的负倒数。

答题过程

展开

The midpoint of ABAB is

(−4+82,11+22)=(2,132).\begin{align*} \left(\frac{-4+8}{2},\frac{11+2}{2}\right) =\left(2,\frac{13}{2}\right). \end{align*}

Since the gradient of ABAB is −34-\frac34, the gradient of ll is 43\frac43.

So

y−132= 43(x−2).\begin{align*} y-\frac{13}{2} =&\,\frac43(x-2). \end{align*}

Multiply by 66:

6y−39= 8x−168x−6y+23= 0.\begin{align*} 6y-39=&\,8x-16\\ 8x-6y+23=&\,0. \end{align*}

(c)

解法一

思路

展开

把 ABAB 当作底边。点 CC 在 ABAB 的垂直平分线上，所以从 CC 到直线 ABAB 的垂直距离就是 MCMC，其中 MM 是 ABAB 的中点。由三角形面积求出 MCMC，再沿着斜率为 43\frac43 的方向移动。斜率 43\frac43 对应的方向向量可以取 (3,4)(3,4)，长度正好是 55。

答题过程

展开

The length of ABAB is

AB= (8−(−4))2+(2−11)2= 122+(−9)2= 15.\begin{align*} AB =&\,\sqrt{(8-(-4))^2+(2-11)^2}\\ =&\,\sqrt{12^2+(-9)^2}\\ =&\,15. \end{align*}

Let MM be the midpoint of ABAB. Since CC lies on the perpendicular bisector, the perpendicular height from CC to ABAB is MCMC.

Using the area,

12(15)(MC)= 37.5MC= 5.\begin{align*} \frac12(15)(MC)=&\,37.5\\ MC=&\,5. \end{align*}

The line ll has gradient 43\frac43, so a direction vector of length 55 along ll is (3,4)(3,4) or (−3,−4)(-3,-4).

Since

M=(2,132),\begin{align*} M=\left(2,\frac{13}{2}\right), \end{align*}

the possible points are

C= (2,132)+(3,4)=(5,212),\begin{align*} C=&\,\left(2,\frac{13}{2}\right)+(3,4) =\left(5,\frac{21}{2}\right), \end{align*}

or

C= (2,132)−(3,4)=(−1,52).\begin{align*} C=&\,\left(2,\frac{13}{2}\right)-(3,4) =\left(-1,\frac52\right). \end{align*}

Therefore

C=(5,212)orC=(−1,52).\begin{align*} C=\left(5,\frac{21}{2}\right) \quad\text{or}\quad C=\left(-1,\frac52\right). \end{align*}