Question

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

Answer 1

Answer:

m : n = 5 : 1

$Point\:of\:Intersection=\left(0,\dfrac{-7}{3}\right)$

Step-by-step explanation:

Given,

Co-ordinates of the point :-

Let,

A = (5 , 6)

B = ( - 1 , - 4)

To Find :-

1. Ration in which 'y'- axis divided the line segment joining the points (5 , 6) and (-1 , -4).
2. Co-ordinates of intersection.

How To Do :-

As we need to find the ratio in which 'y' axis divided those co-ordinates we will get co-ordinates of y-axis = (0 , y). So we got the co-ordinate of y-axis now we need to find the ratio by section formula. To find the point of intersection(0 , y) we need to find the value of 'y' and substitute the value of 'y' in (0 , y) we will get the co-ordinates of point of intersection.

Formula Required :-

Section Formula :-

$(x,y)=\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)$

Solution :-

A = (5 , 6)

Let ,

x₁ = 5 , y₁ = 6

B = (-1 , -4)

Let,

x₂ = - 1  ,  y₂ = - 4

The ratio the y-axis divides the co-ordinates be ' m : n'

According to Question :-

$(0,y)=\left(\dfrac{m(-1)+n(5)}{m+n},\dfrac{m(-4)+n(6)}{m+n}\right)$

$(0.y)=\left(\dfrac{-m+5n}{m+n},\dfrac{-4m+6n}{m+n}\right)$

Equation 'x' co-ordinate to x - co-ordinate and 'y' co-ordinate to y - co-ordinate

$\implies 0=\dfrac{5n-m}{m+n},y=\dfrac{6n-4m}{m+n}$

First Taking '0 = (5n - m)/ (m + n)' :-

$0(m+n)=5n-m$

5n - m = 0

5n = m

m = 5n

m/n = 5/1

∴ m : n = 5 : 1

[ → m = 5 , n = 1]

→ y-axis divides line segment joining the points (5,6) and (-1,-4) in the ration ' 5 : 1'.

Now taking :- y =( 6n - 4m)/(m + n)

$y = \dfrac{6(1)-4(5)}{5+1}$

$=\dfrac{6-20}{6}$

= - 14/6

= - 7/3

∴ y = - 7/3

∴ Point of Intersection = (0 , y) = (0 , - 7/3)