Question

if P(1,-2) is a point on
the line
segment A( x,y) and B (-2,4) such as
that AP:PB= 2:3 find the coordinates
of A​

Answer 1

We're given,

$\longrightarrow \sf{A(x, y) = ({x}_{1}, {y}_{1})}$

$\longrightarrow \sf{P(1, -2) = (\alpha, \beta)}$

$\longrightarrow \sf{B(-2, 4) = ({x}_{2}, {y}_{2})}$

$\longrightarrow \sf{AP:PB = m:n = 2:3}$

$\sf{\\ \\}$

We know by section formula, when a point $\sf{P(\alpha, \beta)}$ intersects internally a line segment AB where $\sf{A({x}_{1},{y}_{1})}$ and $\sf{B({x}_{2},{y}_{2})}$ in the ratio m : n, then:-

$\quad \quad \longrightarrow \sf{\alpha = \dfrac{m{x}_{2} + n{x}_{1}}{m + n}}$

$\quad \quad \longrightarrow \sf{\beta = \dfrac{m{y}_{2} + n{y}_{1}}{m + n}}$

$\sf{\\ \\}$

So, putting the values,

For alpha,

$\longrightarrow \sf{\alpha = \dfrac{m{x}_{2} + n{x}_{1}}{m + n}}$

$\longrightarrow \sf{1 = \dfrac{(2 * -2) + (3 * x)}{2 + 3}}$

$\longrightarrow \sf{1 = \dfrac{ - 4 + 3x}{5}}$

$\longrightarrow \sf{5 = - 4 + 3x}$

$\longrightarrow \sf{9 = 3x}$

$\longrightarrow \underline{\sf{x = 3}}$

$\sf{\\ \\}$

For beta,

$\longrightarrow \sf{\beta = \dfrac{m{y}_{2} + n{y}_{1}}{m + n}}$

$\longrightarrow \sf{-2 = \dfrac{(2 * 4) + (3 * y)}{2 + 3}}$

$\longrightarrow \sf{-2 = \dfrac{ 8 + 3y}{5}}$

$\longrightarrow \sf{-10 = 8 + 3y}$

$\longrightarrow \sf{-18 = 3y}$

$\longrightarrow \underline{\sf{y = -6}}$

$\sf{\\ \\}$

Hence the answer is,

$\longrightarrow{\underline{\underline{\sf{A = (3,-6)}}}}$