Question

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

Answer 1

Answer:

The ratio is 2:9

Step-by-step explanation:

Given: The line 2x+y-4=0 divides the line segment joining A(2,-2) and B (3,7).

To find : The ratio in which line divides?

Solution :

Let the line AB divides by Point C in a ration k:1

Then, Using section formula

$(x_3,y_3)=\frac{x_1n+x_2m}{m+n},\frac{y_1n+y_2m}{m+n}$ 

Let the Point C is (x,y),  A(2,-2) and B (3,7)

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

So,

$x=(\frac{2+3k}{k+1})$  

$y=(\frac{(-2+7k)}{k+1})$  

Substitute in the equation of line,

$2x+y-4=0$

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

$(\frac{4+6k}{k+1})+(\frac{(-2+7k)}{k+1})-4=0$

$4+6k-2+7k-4(k+1)=0$

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

$-2+9k=0$

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

So, The ratio is 2:9

Answer 2

Answer:

2 : 9.

Step-by-step explanation:

We have a line AB, with A(2, -2), B(3, 7) and is intersected by a line 2x + y - 4 = 0.

We'll first find the values of 'x' and 'y', and substitute it in the equation of the intersecting line.

Let us assume that the line is divided in the ratio k : 1.

Hence, we can use the formula:

$\implies \sf P(x,y) = \left( \ \dfrac{kx_2 + x_1}{k+1} \ , \dfrac{ky_2 + x_1}{k+1} \ \right)$

$\implies \sf P(x,y) = \left( \ \dfrac{k(3)+ 2}{k+1} \ , \dfrac{k(7) + (-2)}{k+1} \ \right)$

$\implies \sf P(x,y) = \left( \ \dfrac{3k+2}{k+1} \ , \dfrac{7k-2}{k+1} \ \right)$

Equating 'x' with (3k + 2)/k + 1 and 'y' with (7k - 2)/k + 1 we get,

$\implies \sf x = \dfrac{3k+2}{k+1}$      $\longrightarrow\textcircled{\scriptsize1}$

$\implies \sf y = \dfrac{7k-2}{k+1}$      $\longrightarrow\textcircled{\scriptsize2}$

The Line intersecting AB has the equation:

$\sf \Longrightarrow 2x+y-4=0$

Substitute Equations 1 and 2 above.

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

$\Longrightarrow \sf \dfrac{6k+4}{k+1} + \dfrac{7k-2}{k+1} - 4 = 0$

Taking LCM we get,

$\Longrightarrow \sf \dfrac{6k+4}{k+1} + \dfrac{7k-2}{k+1} - \dfrac{4(k+1)}{k+1} = 0$

$\Longrightarrow \sf \dfrac{6k+4}{k+1} + \dfrac{7k-2}{k+1} - \dfrac{4k-4}{k+1} = 0$

$\Longrightarrow \sf \dfrac{6k+4+7k-2-4k-4}{k+1} = 0$

$\Longrightarrow \sf \dfrac{9k-2}{k+1} = 0$

$\Longrightarrow \sf 9k-2 = 0$

$\Longrightarrow \sf 9k=2$

$\Longrightarrow \sf k=\dfrac{2}{9}$

$\Longrightarrow \sf k:1 =\dfrac{2}{9}:1$

$\Longrightarrow \sf k:1 =2:9$

Hence the line is divided in the ratio 2:9.