# Thread: Find the end point

1. ## Find the end point

Hello guys, kindly help me in this problem. Thank you ^^

The segment connecting (3,0) and (4,-1) is extended each way a distance (7/4) each own length. Find endpoint.

2. Originally Posted by jasonlewiz
Hello guys, kindly help me in this problem. Thank you ^^

The segment connecting (3,0) and (4,-1) is extended each way a distance (7/4) each own length. Find endpoint.
The segment from (3, 0) to (4, -1) has "x range" 4- 2= 1 and "y range" -1- 0= -1: that is, to go from (3, 0) to (4, -1) you go to the right 1 and down 1. The "slope" is -1/1= -1. If you extend that line past (4, -1) you have to keep that same slope so you extend to the right some distance x and down that same distance: y= -x. Of course the distance is given by $\sqrt{x^2+ y^2}= 7/4$ and, since y= -x, that is $\sqrt{x^2+ x^2}= \sqrt{2}x= 7/4$. Add $x= 7/(4\sqrt{2})= 7\sqrt{2}/8$ to 4 and subtract it from 1 to get the (x,y) coordinates of the endpoint in that direction. Go the other way, past (3,0) to get the other endpoint.

3. Hello jasonlewiz
Originally Posted by jasonlewiz
Hello guys, kindly help me in this problem. Thank you ^^

The segment connecting (3,0) and (4,-1) is extended each way a distance (7/4) each own length. Find endpoint.
I assume that you want the coordinates of both end points.

Draw a careful diagram showing these two points on a squared grid. Draw the line segment that joins them and extend it in each direction by $\tfrac74=1\tfrac34$ of its own length. So at the left-hand end that's $1\tfrac34$ to the left and $1\tfrac34$ units up; at the right-hand end that's $1\tfrac34$ units to the right and $1\tfrac34$ units down. If you draw this with care, you should see immediately what the coordinates are.

The left-hand end point has $x$-coordinate $3 - 1\tfrac34=1\tfrac14$; and $y$-coordinate $1\tfrac34$.

The right-hand end has $x$-coordinate $4 + 1\tfrac34 = 5\tfrac34$; and $y$-coordinate $-1-1\tfrac34=-2\tfrac34$.

The line determined by those two points can be written as $\left\langle {3 + t, - t} \right\rangle ,\,t \in \mathbb{R}$.
If $t=0$ we get $(3,0)$ and $t=1$ gives $(4,-1)$.
To get the two required points let $t=\frac{-7}{4}~\&~t=\frac{11}{4}$.