Question

Find the ratio in which y-axis divides the line segment joining the points A(5,-
6) and B(-1,-4).
​

Answer 1

Given :-

• coordinates of point A are (5,-6)
• coordinates of point B are (-1,-4)

To find :-

• The ratio in which y axis divides the line segment joining points A and B .

Knowledge required :-

• Section formula

Section formula tells us the coordinates of point P (x,y) that  Divides the Line segment joining points A (x₁ , y₁) and B (x₂ , y₂) in the ratio m : n .

$\boxed{\sf{(x,y)=\left(\frac{mx_2+nx_1}{m+n}\;\;,\;\;\frac{my_2+my_1}{m+n}\right)}}$

Figure required :-

$\setlength{\unitlength}{1.5cm}\thicklines\begin{picture}(6,4)\put(4,3){\line(1,0){4}}\put(4,2.6){ \sf{A\;(5,-6)} }\put(7.8,2.6){ \sf{B\;(-1,-4)} }\put(5.5,3){\circle*{0.1}}\put(5.5,2.7){ \sf{P} }\put(4.5,3.3){ {\sf{k\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; : \;\;\;\;\;\;\;\;\;\;\;\;\;1} }}\put(5.5,2.4){ \sf{(0,y)} }\end{picture}$

↣ where P (x,y) is the point of intersection of y axis .

↣ We are taking m/n = k/1 and for convenience in solving we are taking AP : BP = k : 1 .  

↣ Since equation of y - axis is given by x = 0 therefore , coordinates of Point P will be ( 0 , y ).

Solution :-

Using section formula

where ,

x = 0

x₁ = 5 , y₁ = -6

x₂ = -1 , y₂ = -4

and  m : n = k : 1

$\implies\sf{(0,y)=\left(\frac{(k)(-1)+(1)(5)}{k+1}\;\;,\;\;\frac{(k)(-4)+(1)(-6)}{k+1}\right)}\\\\\implies\sf{(0,y)=\left(\frac{-k+5}{k+1}\;\;,\;\;\frac{-4k-6}{k+1}\right)}$

that implies

$\implies\sf{0=\frac{-k+5}{k+1}}\\\\\implies\sf{0=-k+5}\\\\\implies\boxed{\sf{k=5}}$

so,

$\rightarrowtail\;\;\sf{m:n=k:1}\\\\\rightarrowtail\boxed{\boxed{\sf{m:n=5:1}}}$  

  Ans.

Hence ,

Line AB will be divided in the ratio 5 : 1  by the y axis .