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I am having difficulty with the following problem

In the given figure the point on segment PQ is twice as from P as from Q is. What is the point? Ans is $(2,1)$. How did that answer get calculated any suggestions

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5 Answers

up vote 1 down vote accepted

There is really no need to use any quadratics or roots.

Hint: Consider the same problem on the plain number line first.

How do you find the number between $2$ and $5$ which is twice as far from $2$ as from $5$?

You take their difference, which is $3$. Now splitting this distance by ratio $2:1$ means the first distance is two thirds, the second is one third, so we get

$$ 4 = 2 + \frac{2}{3}(5-2)$$

It works completely the same with geometric points (using vector operations), just linear interpolation: Call the result $R$, then

$$ R = P + \frac{2}{3}(Q-P)$$ so in your case we get $$ R= (0,-1) + \frac 2 3 (3,3) = (2,1)$$

Why does this work for 2D-distances as well, even if there seem to be roots involved? Because vector length behaves linearly after all! (meaning $|t \cdot \vec a|=t|\vec a|$ for any positive scalar $t$)

Edit: We'll try to divide a distance $s$ into parts $a$ and $b$ such that $a$ is twice as long as $b$. So it's $a = 2b$ and we get $$s = a + b = 2b + b = 3b$$ $$\Leftrightarrow b = \frac 1 3 s \Rightarrow a = \frac 2 3 s$$ That's where the fractions come from.

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The last equality should read with $|t|$. –  user21436 Aug 28 '12 at 9:03
    
Your method does seem interesting and simple –  MistyD Aug 28 '12 at 9:28
    
You said .. "means the first distance is two thirds, the second is one third" could you please explain that, –  MistyD Aug 28 '12 at 9:35
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If the required point is $(x,y)$, then because we're looking at a straight line, $x$ must be twice as far from the x-coordinate of $P$ as it is from the x-coordinate of $Q$. That is, it's twice as far from 0 as it is from 3. Since it's in $[0,3]$, we must have $x=2$. A similar argument for $y$ gives $y=1$.

No distance formula, square roots or quadratic equations are required.

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Hint:

1) Find the equation of the line.

The equation of the line is $y=x-1$.

2) Write your condition by making use of distance formula. Use $1$.

You need: $(x,x-1)$ such that $$2\sqrt{(3-x)^2+(2-(x-1))^2}=\sqrt{(x-0)^2+(x-1-(-1))^2}$$

You get a quadratic. Solving which you have $x=2,6$. But, $x \in [0,3] \Rightarrow x=2$.

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So I get $y=\frac{2x}{3} -1$ How would I add a condition using this equation in the distance formula ? –  MistyD Aug 28 '12 at 8:29
    
Dear Downvoter, can you please explain? –  user21436 Aug 28 '12 at 8:30
    
@MistyD Since a point we are searching for lies on the line, it should be of the form $(x, \frac{2x}{3}-1)$. Now you have distance constraint. Write it out. –  user21436 Aug 28 '12 at 8:33
    
@MistyD I have updated my answer. The equation of your line was wrong. –  user21436 Aug 28 '12 at 8:48
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@wentaway I am NOT the downvoter; but I gave serious consideration to downvoting both your solution and PooyaM's. You are both nuking mosquitoes. Distance formula and quadratics? To divide a baby little line segment into equal thirds? You can do better! –  user22805 Aug 28 '12 at 22:01
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Consider that point has coordinates $H(x_H,y_H)$
Distance "$d$" between $(x_A,y_A)$ and $(x_B,y_B)$ is:
$d_{AB}=\sqrt{(x_A-x_B)^2+(y_A-y_B)^2}$

I:From the questions constraints we have:

$\sqrt{(x_H-0)^2+(y_H-(-1))^2} = 2\times \sqrt{(x_H-3)^2+(y_H-2)^2} $
$x_H^2+(y_H+1)^2=4\times (x_H-3)^2 +4\times(y_H-2)^2$

II:$(x_H,y_H)$ is on the same line connecting P to Q:
So $Slope_{HP}=Slope_{QH}$
$Slope_{AB}=(y_B - y_A)/(x_B-x_A)$
So: $(y_H-(-1)) /(x_H-0) =(2-y_H)/(3-x_H) $

Then we should solve for x and y from the 2 variable 2 equations we derived:
From II : $x_H=y_H+1$
plugging this into I yields: $x_H^2=4(x_H-3)^2$ two answers:
$x_H=2x_H-6 \Rightarrow x_H=6,y_H=5 \chi$ Because It is not between P and Q
$x_H=-2x_H+6 \Rightarrow x_H=2, y_H=1 \checkmark$

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Awesome solution clear and concise. Thanks @PooyaM –  MistyD Aug 28 '12 at 8:46
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The parametric equation of the line joinining $P$ and $Q$ is $$ \vec{OR}=\vec{OP}+t(\vec{OQ}-\vec{OP})=(0,-1)+t(3,3)=(3t,3t-1) $$ where $R$ is its generic point. The parameter $t$ is not the euclidean distance, yet it is a linear reparametrization of the euclidean distance on the line. Thus, since $R=P$ when $t=0$ and $R=Q$ when $t=1$, the point in question is that for which $$ |t|=|2(t-1)| \quad\text{and}\quad 0<t<1, $$ i.e. $t=2/3$, meaning $R=(2,1)$.

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