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IAL 2022 Jan Q8

A Level / Edexcel / P1

IAL 2022 Jan Paper · Question 8

题目

Problem

The line l1l_1 has equation

2x−5y+7=0.\begin{align*} 2x-5y+7=0. \end{align*}

(a) Find the gradient of l1l_1.

(1)

Given that

  • the point AA has coordinates (6,−2)(6,-2)
  • the line l2l_2 passes through AA and is perpendicular to l1l_1

(b) find the equation of l2l_2 giving your answer in the form y=mx+cy=mx+c, where mm and cc are constants to be found.

(3)

The lines l1l_1 and l2l_2 intersect at the point MM.

(c) Using algebra and showing all your working, find the coordinates of MM.

(Solutions relying on calculator technology are not acceptable.)

(3)

Given that the diagonals of a square ABCDABCD meet at MM,

(d) find the coordinates of the point CC.

(2)

解答

(a)

解法一

思路

展开

把直线方程整理成 y=mx+cy=mx+c，其中 mm 就是梯度。

答题过程

展开 2x−5y+7= 0−5y= −2x−7y= 25x+75.\begin{align*} 2x-5y+7=&\,0\\ -5y=&\,-2x-7\\ y=&\,\frac25x+\frac75. \end{align*}

Therefore the gradient of l1l_1 is

25.\begin{align*} \frac25. \end{align*}

(b)

解法一

思路

展开

垂直直线的斜率乘积为 −1-1。所以 l2l_2 的斜率是 −52-\frac52。再用点 A(6,−2)A(6,-2) 代入点斜式。

答题过程

展开

Since l2l_2 is perpendicular to l1l_1,

m2=−52.\begin{align*} m_2=-\frac52. \end{align*}

Using the point A(6,−2)A(6,-2),

y−(−2)= −52(x−6)y+2= −52x+15y= −52x+13.\begin{align*} y-(-2)=&\,-\frac52(x-6)\\ y+2=&\,-\frac52x+15\\ y=&\,-\frac52x+13. \end{align*}

Therefore

l2:y=−52x+13.\begin{align*} l_2:\quad y=-\frac52x+13. \end{align*}

(c)

解法一

思路

展开

交点 MM 同时在两条直线上。把两条直线都写成 y=y= 的形式后令它们相等，先求 xx，再代回求 yy。

答题过程

展开

From l1l_1,

y=25x+75.\begin{align*} y=\frac25x+\frac75. \end{align*}

From l2l_2,

y=−52x+13.\begin{align*} y=-\frac52x+13. \end{align*}

At MM, the two yy values are equal:

25x+75= −52x+1325x+52x= 13−75410x+2510x= 655−752910x= 585x= 4.\begin{align*} \frac25x+\frac75=&\,-\frac52x+13\\ \frac25x+\frac52x=&\,13-\frac75\\ \frac{4}{10}x+\frac{25}{10}x=&\,\frac{65}{5}-\frac75\\ \frac{29}{10}x=&\,\frac{58}{5}\\ x=&\,4. \end{align*}

Substitute x=4x=4 into l1l_1:

y= 25(4)+75= 85+75= 3.\begin{align*} y =&\,\frac25(4)+\frac75\\ =&\,\frac85+\frac75\\ =&\,3. \end{align*}

Therefore

M=(4,3).\begin{align*} M=(4,3). \end{align*}

(d)

解法一

思路

展开

正方形的对角线互相平分，所以 MM 是 ACAC 的中点。已知 A(6,−2)A(6,-2) 和中点 M(4,3)M(4,3)，可用中点公式反求 CC。

答题过程

展开

Let C=(x,y)C=(x,y). Since MM is the midpoint of ACAC,

(6+x2,−2+y2)=(4,3).\begin{align*} \left(\frac{6+x}{2},\frac{-2+y}{2}\right)=(4,3). \end{align*}

So

6+x2= 4,−2+y2= 3.\begin{align*} \frac{6+x}{2}=&\,4,\\ \frac{-2+y}{2}=&\,3. \end{align*}

Hence

6+x= 8,−2+y= 6,\begin{align*} 6+x=&\,8,\\ -2+y=&\,6, \end{align*}

giving

x=2,y=8.\begin{align*} x=2,\quad y=8. \end{align*}

Therefore

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