Question

Point A is located at (5, 6) and point B is located at (8, −2) . What are the coordinates of the point that partitions the directed line segment AB¯¯¯¯¯ in a 1:3 ratio?

Answer 1

Answer:

$(6;3.33)$

Step-by-step explanation:

To find the coordinates of the point that partitions in a 1:3 ratio, we use:

$x=x_{1}+k(x_{2}-x_{1})\\y=y_{1}+k(y_{2}-y_{1})$

Where $k$ is the ration of partitions, $\frac{1}{3}$ in this case.

Now, we replace all values:

$x=x_{1}+k(x_{2}-x_{1})\\x=5+\frac{1}{3}(8-5)=5+\frac{1}{3}3=6$

So, the horizontal coordinate is 6.

$y=y_{1}+k(y_{2}-y_{1})\\y=6+\frac{1}{3}(-2-6)=6+\frac{1}{3}(-8)=3.33$

The vertical coordinate is 3.33.

Therefore, the coordinates of the point that partitions the directed line segment AB in a 1:3 ratio is $(6;3.33)$

Answer 2

Answer:

(6, 3 1/3)

Step-by-step explanation: Split this problem into x and y parts. In x direction, the length of the AB is (8-5)=3. So 1/3 from A would be 5+1/3(8-5)=6

In y, AB goes from 6 to -2, so the coordinate changes by (-2-6)=-8 units.

1/3 along this ling will be 6 + 1/3 (-2-6)=6-2 2/3=3 1/3