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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?
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Answer :

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
kate200468
[tex]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}= \frac{2}{5} \cdot \overrightarrow {AB}=\frac{2}{5}\cdot[-5;\ 5]=[\frac{2}{5}\cdot(-5);\ \frac{2}{5}\cdot5]=[-2;\ 2][/tex]

[tex]\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.[/tex]
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