Question

find the ratio in which the line segments joining the points A(1,-5) and B ( -4 , 5 ) is divided by the z axis find the co ordinates of point of diverse... ​

Answer 1

Answer

Let, M device the line joining of point A(1, -5) and B(1-5) in ratio K:1

By Section formula , we get

coordinates of M

$↪( \frac{ - 4k + 1}{k + 1} ,\frac{5k - 1}{k + 1} ) \: \: \: \: \: \: \: \: \: \: \: (i) \: ....$

According to the question, line segment joining A(1,-5) and B(-4,5) is divided by x axis point and lies on x axis

↪coordinates of M = (X, 0) (ii)....

From equation 1 and 2 , we get

$↪(x,0) = ( \frac{ - 4k + 1}{k + 1} , \frac{5k - 5}{k - 1} ) \: \: \: \: \: \: \: \:( iii)...$

On comparing y-coordinate both side , we get

$↪ \frac{5k - 5}{k + 1} = 0$

$↪5k - 5 = 0$

$↪k = 1$

On putting value of k in equation( iii) , we get

$↪(x,0) = ( \frac{ - 4 \times 1 + 1}{1 + 1} , \frac{5 \times 1}{1 + 1} )$

$↪( \frac{ - 4 + 1}{2} , \frac{5 - 5}{2} )$

Therefore,

$↪(x,0) = ( \frac{ - 3}{2} ,0)$

So the required ratio is 1:1 and point of division M is

$( \frac{ - 3}{2} ,1)$

Answer 2

Let, M device the line joining of point A(1, -5) and B(1-5) in ratio K:1  

By Section formula , we get

coordinates of M

$↪( \frac{ - 4k + 1}{k + 1} ,\frac{5k - 1}{k + 1} ) \: \: \: \: \: \: \: \:(i) …$

According to the question, line segment joining A(1,-5) and B(-4,5) is divided by x axis point and lies on x axis

↪coordinates of M = (X, 0) (ii)....

From equation 1 and 2 , we get

$↪(x,0)( \frac{ - 4k + 1}{k + 1} ,\frac{5k -1 }{k - 1} ) \: \: \: \: \: \: \: \:(iii) …$

On comparing y-coordinate both side , we get

$↪( \frac{5k -5 }{k + 1} = 0)$

$↪({5k -5 } = 0)$

$↪( k =1)$

$↪(x,0)=( \frac{4*1+1 }{1+1}, \frac{5*1}{1+1})$

$↪( \frac{-4+1}{2}, \frac{5-5}{2})$

Therefore,

$↪(x,0)=( \frac{-3 }{2}, 0)$

So the required ratio is 1:1 and point of division M is

$( \frac{-3 }{2}, 1)$