# 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. - Mathematics

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.

#### Solution 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

#### Solution 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.

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#### APPEARS IN

RD Sharma Class 10 Maths
Chapter 6 Co-Ordinate Geometry
Exercise 6.3 | Q 21 | Page 29
RS Aggarwal Secondary School Class 10 Maths
Chapter 16 Coordinate Geomentry
Q 32