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
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
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
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.
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, $x \in [0,3] \Rightarrow x=2$.
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
$x_H=-2x_H+6 \Rightarrow x_H=2, y_H=1 \checkmark$
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.