A line passes through A(1,1) , B(4,3). Find the coordinates of the point on the line which is (1) three times as far from A as from B, (2) three times as far from B as from A.
Treat the coordinates, x from 1 to 4 and y from 1 to 3, separately. "Three times as far from A as from B". Think of that as "3" (distance from A) and "1" (distance from B). 3+ 1= 4 so divide the distance from 1 to 4 into 4 separate parts. 4-1= 3 so each part has x-length 3/4. The x-distance from A to the new point is 3(3/4)= 9/4 so the x-coordinate is 1+ 9/4= 13/4 or 3 and 1/4. Add another 3/4 to that and we get 4, the x-coordinate of B. The 3- 1= 2 so each part has y-length 2/4= 1/3. 3- 1/2= 5/2= 2 and 1/2 (subtract from because the point we want is between A and B). The y-coordinate of the point is 1/2 and the point itself is (9/4, 5/2).
I said "the point we want is between A and B" because the problem specifiacally said "the point on the line. Actually, there is another point on the line through A and B and three times as from from A as from B. It is a point beyond B. 3- 1= 2 so divide the segement from A to B in half. The midpoint is ((1+ 4)/2, (1+ 3)/2)= (5/2, 2), an x-distance of 5/2- 1= 3/2 and a y-distance of 2- 1= 1. Multiplying each of those by 3, (1, 1)+ (9/2, 3)= (11/2, 4) lies on the line through A and B. The distance from A to that point is $\displaystyle \sqrt{(11/2- 1)^2+ (4- 1)^2}= \sqrt{\frac{81}{4}+ 9}= \sqrt{\frac{117}{4}}= 3\sqrt{13}/2$ while the distance from B to that point is $\displaystyle \sqrt{(4- 11/2)^2+ (3- 4)^2}= \sqrt{9/4+ 1}= \sqrt{\frac{13}{4}}= \sqrt{13}/2$.
Hello, aeroflix!
Here is a primitive solution . . .
A line passes through $\displaystyle A(1,1),\;B(4,3).$
Find the coordinates of the point on the line which is:
(1) three times as far from A as from B
(2) three times as far from B as from A.Code:| (4,3) | o B | P * : | o : | * : | * : | * : | * : | * : | A o - - - - - - - - - - - + | (1,1) (4,1) | - - + - - - - - - - - - - - - - - - - - |
(a) Going from $\displaystyle \,A$ to $\displaystyle \,B$, we move: .$\displaystyle \begin{Bmatrix}\text{3 units right} \\ \text{2 units up} \end{Bmatrix}$
We want $\displaystyle \,P$ to be $\displaystyle \tfrac{3}{4}$ of the way from $\displaystyle \,A$ to $\displaystyle \,B.$
So we will move: .$\displaystyle \begin{Bmatrix}
\frac{3}{4}(3) &=& \frac{9}{4}\text{ units right} \\ \\[-3mm]
\frac{3}{4}(2) &=& \frac{3}{2}\text{ units up} \end{Bmatrix}\:\text{ from point }A.$
Hence: .$\displaystyle \begin{Bmatrix}x &=& 1 + \frac{9}{4} &=& \frac{13}{4} \\ \\[-4mm] y &=& 1 + \frac{3}{2} &=& \frac{5}{2} \end{Bmatrix} $
Therefore: .$\displaystyle P\left(3\frac{1}{4},\,2\frac{1}{2}\right)$