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​

Respuesta :

Answer:

[tex]P=\left(\dfrac{13}{3},-\dfrac{1}{3}\right)[/tex]

[tex]|\overrightarrow{MN}|=4\sqrt{5}[/tex]

Step-by-step explanation:

Part (a)

[tex]x_P=\dfrac13(x_N-x_M)+x_M=\dfrac13(7-3)+3=\dfrac{13}{3}[/tex]

[tex]y_P=\dfrac13(y_N-y_M)+y_M=\dfrac13(5-(-3))+(-3)=-\dfrac13[/tex]

[tex]\implies P=\left(\dfrac{13}{3},-\dfrac{1}{3}\right)[/tex]

Part (b)

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

Using the distance between two points formula, where

[tex](x_1,y_1)=(3,-3)[/tex]  and  [tex](x_2,y_2)=(7,5)[/tex]

[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]

[tex]\implies |\overrightarrow{MN}|=\sqrt{(7-3)^2+(5-(-3))^2}[/tex]

[tex]\implies |\overrightarrow{MN}|=4\sqrt{5}[/tex]