The following is a very useful fact. The points in the interval from $(a,b)$ to $(c,d)$ are precisely the points $(x,y)$ such that $$x=(1-t)a+ tc,\qquad y=(1-t)b+td,\tag{$\ast$}$$ where $t$ ranges over the interval from $0$ to $1$. Moreover, this point $(x,y)$ divides the interval from $(a,b)$ to $(c,d)$ in the ratio $t:1-t$. To put it another way, the distance from $(a,b)$ to $(x,y)$ is $t$ times the distance from $(a,b)$ to $(c,d)$. We will use $(\ast)$, with $(a,b)=(0,3)$ and $(c,d)=(3,0)$. The distance from $(0,3)$ to $(3,0)$ is $\sqrt{18}$, or more simply $3\sqrt{2}$. If $B$ is to be at distance $d$ from $A$, we want $t:1 =d:3\sqrt{2}$. So $t=\dfrac{d}{3\sqrt{2}}$. Finally, use this value of $t$ in $(\ast)$ to find the coordinates $(x,y)$ of $B$.
Here the point (12,5) is 12 units along, and 5 units up We can use Cartesian Coordinates to locate a point by how far along and how far up it is: And when we know both end points of a line segment we can find the midpoint "M" (try dragging the blue circles): The midpoint is halfway between the two end points:
To calculate it:
As a formula: M = ( xA+xB 2 , yA+yB 2 )
Use the formula: M = ( xA+xB 2 , yA+yB 2 ) M = ( (−3)+8 2 , 5+(−1) 2 ) M = ( 5/2, 4/2 ) M = ( 2.5, 2 ) Copyright © 2017 MathsIsFun.com |