Question

find the ratio in which y_axis divides the line segment joining the points (5,-6)and (-1,-4)also find the point of intersection​

Answer 1

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

Given :-

Points (5,-6) and (-1,-4)

To find:-

Find the ratio in which y_axis divides the line segment joining the points (5,-6)and (-1,-4)also find the point of intersection ?

Solution:-

Given points are (5,-6) and (-1,-4)

Let (x1, y1)=A(5,-6)=>x1=5 and y1=-6

Let(x2, y2)=B(-1,-4)=>x2=-1 and y2=-4

Let the ratio in which y-axis divides the linesegment joining the points be m1:m2

The point on y-axis =P (0,y)

A______P_______________________B

We know that

The coordinates of the point in which divides the linesegment joining the points (x1, y1) and (x2, y2) in the ratio m1:m2 is

( {m1x2+m2x1}/(m1+m2) ,{m1y2+m2y1}/(m1+m2) )

=> P(0,y)

({(m1)(-1)+(m2)(5)}/(m1+m2),{(m1)(-4)+(m2)(-6) / (m1+m2))

=> ( {-m1+5m2}/(m1+m2) , {-4m1-6m2}/(m1+m2) )

On Comparing both sides then

=> {-m1+5m2}/(m1+m2) = 0

=> -m1+5m2 = 0(m1+m2)

=> -m1+5m2 = 0

=> -m1 = -5m2

=> m1 = 5m2

=> m1 / m2 = 5

=> m1/m2 = 5/1

=> m1 : m2 = 5:1

and

{-4m1-6m2}/(m1+m2) = y

=> y = [-4(5)-6(1)]/[5+1]

=> y = (-20-6)/6

=> y = -26/6

=> y = -13/3

Answer:-

The required ratio for the given problem is 5:1

The required intersecting point =

( 0, -13/3 )

Used formulae:-

• The equation of y-axis is x=0

• The coordinates of the point in which divides the linesegment joining the points (x1, y1) and (x2, y2) in the ratio m1:m2 is
• ({m1x2+m2x1}/(m1+m2),{m1y2+m2y1}/(m1+m2) )