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IAL 2025 Oct A Q4

A Level / Edexcel / P1

IAL 2025 Oct A Paper · Question 4

题目

Problem

Figure 2 shows the points PP, QQ and RR.

Figure 2

Points PP and QQ have coordinates (1,4)(-1,4) and (4,7)(4,7) respectively.

(a) Find an equation for the straight line passing through points PP and QQ. Give your answer in the form ax+by+c=0ax+by+c=0 where aa, bb and cc are integers.

(3)

The point RR has coordinates (p,3)(p,-3), where pp is a positive constant.

Given that angle QPR=90QPR=90^\circ,

(b) find the value of pp.

(Solutions relying on calculator technology are not acceptable.)

(3)

解答

(a)

解法一

思路

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先用两点求斜率,再代入点斜式,最后整理到 ax+by+c=0ax+by+c=0

答题过程

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The gradient of PQPQ is

744(1)=35.\begin{align*} \frac{7-4}{4-(-1)} =\frac35. \end{align*}

Using point P(1,4)P(-1,4),

y4=35(x+1)5y20=3x+33x5y+23=0.\begin{align*} y-4=&\,\frac35(x+1)\\ 5y-20=&\,3x+3\\ 3x-5y+23=&\,0. \end{align*}

(b)

解法一

思路

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因为 QPR=90\angle QPR=90^\circ,所以 PRPR 垂直于 PQPQ。先求出 PRPR 的斜率应为 53-\frac53,再用 P(1,4)P(-1,4)R(p,3)R(p,-3) 写斜率方程。

答题过程

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The gradient of PQPQ is 35\dfrac35, so the gradient of PRPR is

53.\begin{align*} -\frac53. \end{align*}

Using P(1,4)P(-1,4) and R(p,3)R(p,-3),

34p(1)=537p+1=53.\begin{align*} \frac{-3-4}{p-(-1)}=&\,-\frac53\\ \frac{-7}{p+1}=&\,-\frac53. \end{align*}

Hence

21=5(p+1)21=5p+55p=16p=165.\begin{align*} 21=&\,5(p+1)\\ 21=&\,5p+5\\ 5p=&\,16\\ p=&\,\frac{16}{5}. \end{align*}