Question

M is the point (3, -3) and N is the point (7,5). P is a point on the line MN such that MP = 1/3MN. - (a) Find the position vector of P. (b) Calculate the magnitude of MN​

Answer 1

`P=\left((13)/(3),-(1)/(3)\right)`

`|\overrightarrow{MN}|=4√(5)`

Explanation:

Part (a)

`x_P=\frac13(x_N-x_M)+x_M=\frac13(7-3)+3=(13)/(3)`

`y_P=\frac13(y_N-y_M)+y_M=\frac13(5-(-3))+(-3)=-\frac13`

`\implies P=\left((13)/(3),-(1)/(3)\right)`

Part (b)

The magnitude of MN is the distance between points M and N.

Using the distance between two points formula, where

`(x_1,y_1)=(3,-3)` and
`(x_2,y_2)=(7,5)`

`d=√((x_2-x_1)^2+(y_2-y_1)^2)`

`\implies |\overrightarrow{MN}|=√((7-3)^2+(5-(-3))^2)`

`\implies |\overrightarrow{MN}|=4√(5)`