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*13}=5\sqrt[]{13}`

`=5*3.606`

`=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

`(2)/(5)*8+2=(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

`(2)/(5)*8+4=(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)