Question

The ratio in which the y-axis divides the line segment joining the points (5, -6) and (-1, - 4) is
(a) 1 : 5
(b) 5 : 1
(c) 2:3
(d) 3:2​

Answer 1

Solution

Let the ratio be K:1

On y-axis the value of x will be 0

So the required point dividing the line joining (5,-6) and (-1,-4) is P(0,y)

Now using section formula

$\dag\sf\ \ P(x,y)=\bigg\lgroup \dfrac{mx_2+nx_1}{m+n}\ ;\ \dfrac{my_2+ny_1}{m+n}\bigg\rgroup$

We have :-

$\sf\ x=0 \ ; \ \ m= k \ ;\ \ x_1= 5\ ;\ \ y_1= (-6)\\ \sf \ \ \ \ y= y\ \ ; \ \ n= 1 \ ;\ \ \: \ x_2= (-1)\ \ ;\ \ y_2=( -4)$

Now,

$:\implies\sf\ \ P(0,y)= \bigg\lgroup\dfrac{k\times (-1)+1\times 5}{k+1}\ ;\ \ \dfrac{k\times (-4)+ 1\times (-6)}{k+1}\bigg\rgroup\\ \\ \\ :\implies\sf\ \ P(0,y)= \bigg\lgroup\dfrac{-k+5}{k+1}\ ;\ \dfrac{-4k-6}{k+1}\bigg\rgroup\\ \\ \\ \sf\ \ By\ equating\ \ them\\ \\ \\ :\implies\sf\ 0= \dfrac{-k+5}{k+1}\\ \\ \\ :\implies\sf\ \ 0(k+1)= -k+5\\ \\ \\ :\implies\sf\ 0= -k+5\\ \\ \\ :\implies\underline{\boxed{\sf\ k=5}}$

$\therefore\underline{\textsf{\textbf{The\ Required\ ratio \ is 5:1}}}$

Option (b) is the correct option