Question

Find point P such that the segment with endpoints A (2,4) and B (17, 14) in a ratio of 2:3P: (___,___)

Answer 1

The given points are A92,4) and B(17,14).

Point P is on the segment AB.

The distance between AB is

$D=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}$

$\text{Substitute }x_2=17,x_1=2,y_2=14\text{ and }y_1=4\text{ as follows:}$

$D=\sqrt[]{(17-2_{})^2+(14_{}-4_{})^2}$

$=\sqrt[]{(15_{})^2+(10)^2}$

$=\sqrt[]{225+100}=\sqrt[]{325}$

$=\sqrt[]{25\times13}=5\sqrt[]{13}$

$=5\times3.606$

$=18.0277$

$D=18$

The distance between A and B is 8.

To find the x-coordinate of the point P compute 2/5 of the distance between A and B and add its value to the x-coordinate of A.

The x-coordinate of P is

$\frac{2}{5}\times8+2=\frac{16}{5}+2$

$=3.2+2$

$=5.2$

To find the y-coordinate of the point P compute 2/5 of the distance between A and B and add its value to the y-coordinate of A.

The y-coordinate of P is

$\frac{2}{5}\times8+4=\frac{16}{5}+4$

$=3.2+4$

$=7.2$

The point P is

$P\colon\text{ ( 5.2, 7.2 )}$

After round the answer, we get point P is (5,7).

Hence the point P is (5,7)