Question

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

• A. 2 : 9
• B. 9 : 2
• C. 9 : 1
• D. 1 : 9

Answer 1

The correct option is A 2 : 9

$\text{Given,}A(2,-2)\&B(3,7).\text{Let point C divides AB in ratio}k:1C(x,y)=(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})C(x,y)=(\frac{3k+2}{k+1},\frac{7k-2}{k+1})\text{The point C lies on the line}2x+y-4=0\therefore 2\left(\frac{3k+2}{k+1}\right)+\left(\frac{7k-2}{k+1}\right)-4=02(3k+2)+(7k-2)=4\times (k+1)6k+4+7k-4k-4-2=0(6+7-4)k+(-2)=0\Rightarrow 9k-2=0k=\frac{2}{9}\therefore k:1=\frac{2}{9}:1=2:9\therefore \text{The required ratio k : 1 is}2:9$