Question

Determine the ratio in which the line $2x+y-4=0$ divides the line segment joining the points A(2, -2) and B (3, 7).

Answer 1

Let the required ratio be k : 1 and let point C divide them in the above ratio.

We know that the coordinates of the point that divides a line segment in the ratio m : n is given by

$(\frac{n\times x_1+m\times x_2}{m+n},\frac{n\times y_1+m\times y_2}{m+n})$

where $(x_1,y_1)and(x_2,y_2)$ are the coordinates of the endpoints of the line segment.

$\therefore$ Coordinates of C are $(\frac{3k+2}{k+1},\frac{7k-2}{k+1})$

Since point C lies on the line represented by the equation
$2x+y-4=0$, it will satisfy the equation.

$\therefore$ So we have

$2\left(\frac{3k+2}{k+1}\right)+\left(\frac{7k-2}{k+1}\right)-4=0$

$\Rightarrow 2(3k+2)+(7k-2)=4\times (k+1)$

$\Rightarrow 6k+4+7k-2-4k-4=0$

$\Rightarrow 9k-2=0$

$\Rightarrow k=\frac{2}{9}$

$\therefore$ The required ratio is k : 1= 2 : 9.