Question

The point P divides the join of (2,1) and (-3, 6) in the ratio  2:3. does P lie on the line x-5y+15=0?

PLEASE ANSWER FAST IT IS KINDA URGENT...

Answer 1

`A=(2;1)\ \ \ and\ \ \ B=(-3;6)\ \ \ \Rightarrow\ \ \ \overrightarrow {AB}=[-3-2;\ 6-1]=[-5;\ 5]\\\\P=(x_P;\ y_P)\ \ \ \Rightarrow\ \ \ \overrightarrow {AP}=[x_P-2;\ y_P-1]\\\\the\ ratio\ \ is\ \ 2:3\\\\.\ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow\ \ \ \overrightarrow {AP}= (2)/(5) \cdot \overrightarrow {AB}=(2)/(5)\cdot[-5;\ 5]=[(2)/(5)\cdot(-5);\ (2)/(5)\cdot5]=[-2;\ 2]`


`\overrightarrow {AP}=[x_P-2;\ y_P-1]\ \ \ \ and\ \ \ \ \overrightarrow {AP}=[-2;\ 2]\\\\.\ \ \ \ \ \ \ \ \ \ \Rightarrow\ \ \ x_P-2=-2\ \ \ \ and\ \ \ \ y_P-1=2\\\\.\ \ \ \ \ \ \ \ \ \ \Rightarrow\ \ \ x_P=0\ \ \ \ \ \ \ \ \ \ \ and\ \ \ \ y_P=3\ \ \ \Rightarrow\ \ \ P=(0;\ 3)\\\\the\ line:\ x-5y+15=0\\\\for\ P=(0;\ 3)\ \ \ \rightarrow\ \ \ 0-5\cdot3+15=-15+15=0\\\\The\ point\ P\ lies\ on\ the\ line.`

Answer 2

x= [ 2*(-3)+ 3*(2)/( 2+3) = 0
y = [ 1*3+ 6*2 ]/2+3 = 3
so P is( 0,3 )
if it lies on the line x+ 5y = 15 then this point should satisfy the equation
so 0 + 5*3 = 15
so Plies on the given line x - 5y + 15 = 0