Question

find the ratio in which P (4, m) divides the line segment joining th points A (2,3) andB (6,-3) .find m

Answer 1

The answer is given below :

Let, the ratio in which AB line is divided by P be a : b.

Then, we have :

A (2, 3) _____a___(P)_____b___ B (6, - 3)

So, the point P is given by

((2b + 6a)/(a + b), (-3a + 3b)/(a + b))

This point is also the point P (4, m).

So, we can write :

(2b + 6a)/(a + b) = 4

=> 2b + 6a = 4a + 4b

=> 2a = 2b

=> a = b

Thus the required ratio is 1 : 1.

So, the point P is

(4, 0) Ξ (4, m)

So, m = 0

Thank you for your question.

Answer 2

Hey friend, Harish here.

Here is your answer:

Given that,

A line segment with points A(2,3) and B (6 , -3)

To find,

The ratios in which the point divides. And the value of m.

Solution,

Let the line be,

                                                       P(4,m)
              A(2,3).__________________|__________________.B(6,-3)


& Let P Divide the line AB in the ratio of k:1

Then By  using section formula,

$\frac{(6\times k)+2}{(k+1)}= 4$

⇒ $(6k + 2) = 4k + 4$

⇒ $2k = 2$

⇒ $k = \frac{2}{2} = 1$
 _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 

We know that,

⇒ $\frac{(k\times (-3))+3}{(k+1) }=m$

Now substituting the K value we get,

⇒ $m = \frac{-3+3}{2} = 0$

Therefore P(4,m) Divides the line in the ratio 1:1.

& The value of m is 0.
______________________________________________

Hope my answer is helpful to you.