Question

Find the ratio in which the point (-1, y) lying on the line segment joining points A(-3, 10) and (6, -8) divides it. Also, find the value of y.

Answer 1

Let k be the ratio in which  P(-1,y ) divides the line segment joining the points

A(-3,10) and B (6,-8) 

Then , 

`(-1,y ) = ((k(6) -3)/(k+1) , (k(-8)+10)/(k+1) )`

`⇒(k(6) -3 )/(k+1) = -1 and y = (k(-8)+10)/(k+1)`

`⇒ k = 2/7`

`"Substituting " k=2/7 "in" y = (k(-8)+10)/(k+1) `, we get

`y =((-8xx2)/(7)+10)/(2/7 +1) = (-16+70)/9 = 6`

Hence, the required ratio is 2 : 7 and y=6

Answer 2

Suppose P(−1, y) divides the line segment joining A(−3, 10) and B(6 −8) in the ratio k : 1.
Using section formula, we get
Coordinates of P = \[\left( \frac{6k - 3}{k + 1}, \frac{- 8k + 10}{k + 1} \right)\]

\[\therefore \left( \frac{6k - 3}{k + 1}, \frac{- 8k + 10}{k + 1} \right) = \left( - 1, y \right)\]

\[\Rightarrow \frac{6k - 3}{k + 1} = - 1\] and \[y = \frac{- 8k + 10}{k + 1}\]

Now,

\[\frac{6k - 3}{k + 1} = - 1\]
\[ \Rightarrow 6k - 3 = - k - 1\]
\[ \Rightarrow 7k = 2\]
\[ \Rightarrow k = \frac{2}{7}\]

So, P divides the line segment AB in the ratio 2 : 7.
Putting k = \[\frac{2}{7}\]  in  \[y = \frac{- 8k + 10}{k + 1}\] , we get

\[y = \frac{- 8 \times \frac{2}{7} + 10}{\frac{2}{7} + 1} = \frac{- 16 + 70}{2 + 7} = \frac{54}{9} = 6\]

Hence, the value of y is 6.