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IAL 2025 May Q5

A Level / Edexcel / P1

IAL 2025 May Paper · Question 5

题目

Problem

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

The line l1l_1 has equation

x−2y+25=0.\begin{align*} x-2y+25=0. \end{align*}

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

(a) Find an equation for l2l_2.

(2)

The lines l1l_1 and l2l_2 intersect at the point PP.

(b) Use algebra to find the coordinates of PP.

(3)

(c) Hence find the shortest distance from l1l_1 to the origin. Write your answer as a fully simplified surd.

(2)

解答

(a)

解法一

思路

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先把 l1l_1 写成 y=mx+cy=mx+c，找出斜率。垂直直线的斜率是负倒数。因为 l2l_2 过原点，所以截距为 00。

答题过程

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For l1l_1,

x−2y+25= 02y= x+25y= 12x+252.\begin{align*} x-2y+25=&\,0\\ 2y=&\,x+25\\ y=&\,\frac12x+\frac{25}{2}. \end{align*}

So the gradient of l1l_1 is 12\frac12.

The gradient of a perpendicular line is

−2.\begin{align*} -2. \end{align*}

Since l2l_2 passes through the origin,

y=−2x.\begin{align*} y=-2x. \end{align*}

(b)

解法一

思路

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交点同时满足两条直线方程。把 y=−2xy=-2x 代入 l1l_1 即可。

答题过程

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Substitute y=−2xy=-2x into x−2y+25=0x-2y+25=0:

x−2(−2x)+25= 05x+25= 0x= −5.\begin{align*} x-2(-2x)+25=&\,0\\ 5x+25=&\,0\\ x=&\,-5. \end{align*}

Then

y=−2(−5)=10.\begin{align*} y=-2(-5)=10. \end{align*}

So

P=(−5,10).\begin{align*} P=(-5,10). \end{align*}

(c)

解法一

思路

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从原点到直线的最短距离一定沿垂线，所以就是 OPOP 的长度。

答题过程

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The shortest distance from l1l_1 to the origin is OPOP.

OP= (−5)2+102= 25+100= 125= 55.\begin{align*} OP =&\,\sqrt{(-5)^2+10^2}\\ =&\,\sqrt{25+100}\\ =&\,\sqrt{125}\\ =&\,5\sqrt5. \end{align*}