Question

Find the ratio in which the point P(–1, y) lying on the line segment joining points A(–3, 10) and B(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.