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

STEP 1: Finding the coordinates of the point

Let the point $C(x,y)$ divides the line segment joining the points $A(2,-2)$ and $B(3,7)$ in the ratio $k:1$

This point $C(x,y)$ is common for both line $2x+y-4=0$ and line joining the points $A(2,-2)$ and $B(3,7)$.

By section formula,

$(x,y)=\left(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n}\right)$

Therefore,

$C(x,y)=\left(\frac{k\times 3+1\times 2}{k+1},\frac{k\times 7+1\times (-2)}{k+1}\right)$

$C(x,y)=\left(\frac{3k+2}{k+1},\frac{7k-2}{k+1}\right)$

$\Rightarrow x=\frac{3k+2}{k+1}andy=\frac{7k-2}{k+1}$

STEP 2 : Substituting the values of $x$ and $y$ in the equation of line

We substitute the values of $x$ and $y$ in the equation of line we get

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

$\frac{6k+4+7k-2}{k+1}=4$

$13k+2=4k+4$

$13k-4k=4-2$

$9k=2$

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

Therefore, the ratio in which the line $2x+y-4=0$ divides the line segment joining the points $A(2,-2)$ and $B(3,7)$ is $2:9$ internally.