Question

the fault line 3x+y-9=0 divides the line joining the country p( 1 , 3 ) and country q ( 2 , 7 ) internally in the ratio​

Answer 1

Step-by-step explanation:

Determine the ratio in which the line 3x + y - 9 = 0 divides the segment joining the points (1, 3) and (2, 7). Let the given points be A( 1, 3) and B(2, 7). So, the required ratio be (3 : 4).

Answer 2

Given:

The fault line 3x + y - 9 = 0

p( 1 , 3 ) and q( 2 , 7 )

To find :

The ratio​ in which the fault line 3 x + y - 9 = 0 divides the line joining the country p ( 1 , 3 ) and country q ( 2 , 7 ) internally.

Formula to be used:

(x , y) = $(\frac{mx_2+ nx_1}{m+n} , \frac{my_2+ ny_1}{m+n} )$

Solution:

Step 1 of 2:

The fault line 3 x + y - 9 = 0 divides the line joining the country P ( 1 , 3 ) and country Q ( 2 , 7 ) internally.

Therefore the ratio of m:n will be k:1

Substitute

m = k , n = 1

$x_1 = 1 ,y_1 = 3\\x_2 = 2, x_2 = 7$   in the following formula,

(x , y) = $(\frac{mx_2+ nx_1}{m+n} , \frac{my_2+ ny_1}{m+n} )$

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

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

Therefore,

x = $\frac{2k+ 1}{k+1}$

y = $\frac{7k+ 3}{k+1}$

Step 2 of 2:

Substituting the values of x and y in the following equation

3 x + y - 9 = 0

$3(\frac{2k+ 1}{k+1})$ +  $\frac{7k+ 3}{k+1}$ - 9 = 0

$\frac{6k+ 3}{k+1}$ +  $\frac{7k+ 3}{k+1}$ - 9 = 0

$\frac{6k+ 3+7k+3}{k+1}$ - 9 = 0

$\frac{13k+ 6}{k+1}$ - 9 = 0  

$\frac{13k+ 6-9(k+1)}{k+1}$ = 0

$\frac{13k+ 6-9k-9}{k+1}$ = 0

$\frac{4k-3}{k+1}$ = 0

4k - 3 = 0(k -1)

4k - 3 = 0

4k = 3

k = $\frac{3}{4}$

The required ratio is 3 : 4

Final answer:

The fault line 3x + y - 9 = 0 divides the line joining the country p( 1 , 3 ) and country q( 2 , 7 ) internally in the ratio​ 3 : 4 .