# Find the ratio in which y-axis divides the line segment joining the points A(5, –6) and B(–1, –4). Also find the coordinates of the point of division. - Mathematics

Sum

Find the ratio in which y-axis divides the line segment joining the points A(5, –6) and B(–1, –4). Also find the coordinates of the point of division.

#### Solution 1

Let (0, α) be a point on the y-axis dividing the line segment AB in the ratio k : 1.

Now, using the section formula, we get

(0,alpha)=((-"k"+5)/("k"+1),(-4"k"-6)/("k"+1))

=>(-"k"+5)/("k"+1)=0,(-4"k"-6)/("k"+1)=alpha

Now,

(-"k"+5)/("k"+1)=0

⇒ k + = 0

⇒ 5

Also,

(-4"k"-6)/("k"+1)=alpha

=>(-4xx5-6)/(5 +1)=alpha

=>alpha=(-26)/6

=>alpha=-13/3

Thus, the y-axis divides the line segment in the ratio k : 1, i.e. 5 : 1.

Also, the coordinates of the point of division are (0, α), i.e (0,-13/3)

#### Solution 2

The ratio in which the y-axis divides two points (x1 , y1)  and  (x2 , y2)  is  $\lambda: 1$

The co-ordinates of the point dividing two points (x1 , y1)  and (x2 , y2)   in the ratio m : n  is given as,

(x , y) = ((lambdax_2 + x_1)/(lambda + 1 )) ,((lambda"y"_2 + "y"_1)/(lamda + 1))  where, lambda = "m"/"n"

Here the two given points are A(5,−6) and B(−1,−4).

(x, "y") = ((-lambda + 5)/(lambda + 1),(- 4lambda - 6)/(lambda + 1))

Since, the y-axis divided the given line, so the x coordinate will be 0.

$\frac{- \lambda + 5}{\lambda + 1} = 0$

$\lambda = \frac{5}{1}$

Thus the given points are divided by the y-axis in the ratio  5 : 1.

The co-ordinates of this point (x, y) can be found by using the earlier mentioned formula.

(x , "y" ) = ((5/1 (-1) + (5) )/(5/1 + 1)) , ((5/1(-4)+(-6))/(5/1 +1))

(x , "y") = (0/6) , (-26/6)

(x , "y")  = ( 0 , - 26/6)

Thus the co-ordinates of the point which divides the given points in the required ratio are (0,-26/6).

#### Notes

Students should refer to the answer according to their question and marks.

Concept: Section Formula
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Chapter 6: Co-Ordinate Geometry - Exercise 6.3 [Page 29]

#### APPEARS IN

RD Sharma Class 10 Maths
Chapter 6 Co-Ordinate Geometry
Exercise 6.3 | Q 26 | Page 29

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