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).

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

Answer 1

The correct option is A

2 : 9

Let the required ratio be k : 1 and the point C divide them in the above ratio,
$\therefore$ Coordinates of C will be $(\frac{3k+2}{k+1},\frac{7k-2}{k+1})$
Since the point C lies on the given line, $2x+y-4=0$
$\therefore$ We have $2\left[\left(\frac{3k+2}{k+1}\right)\right]+\left(\frac{7k-2}{k+1}\right)-4=0$
$\Rightarrow 2(3k+2)+(7k-2)=4\times (k+1)\Rightarrow 6k+4+7k-4k-4-2=0\Rightarrow (6+7-4)k+(-2)=0\Rightarrow 9k-2=0\Rightarrow k=\frac{2}{9}$
$\therefore$ The required ratio =$k:1=\frac{2}{9}:1=2:9$