Question

In what ratio are the joining points $(3, - 6)$ and $( - 6,8)$ divided by the $y$- axis?

Answer 1

In this question we will use the section formula to solve this question. We know that if point$(x,y)$ divides the line joining the points $({x_1},{y_1})$ and $({x_2},{y_2})$ in the ratio $m:n$, then we can say that the point of division of the coordinates are $(x,y) = \left( {\dfrac{{m{x_2} + n{x_1}}}{{m + n}},\dfrac{{m{y_2} + n{y_1}}}{{m + n}}} \right)$. Using this we can solve the given question.

Complete step by step answer:
First let us assume the ratio that the $y -$axis that divides the lines joining the points $(3, - 6)$ and $( - 6,8)$ be $k:1$ and the point of intersection of this line to the y- axis to be $(0,y)$.
We will now use the section formula i.e. $(x,y) = \left( {\dfrac{{m{x_2} + n{x_1}}}{{m + n}},\dfrac{{m{y_2} + n{y_1}}}{{m + n}}} \right)$.
With comparing from the question we have
${x_1} = 3,{y_1} = - 6,m = k,n = 1$ and ${x_2} = - 6,{y_2} = 8$.
We know that the $x$ coordinate of the point of division is $0$. So using this we can write: $0 = \dfrac{{ - 6 \times k + 3 \times 1}}{{k + 1}}$.
On solving we have
$0 = \dfrac{{ - 6k + 3}}{{k + 1}} \Rightarrow 0 = - 6k + 3$.
By transferring the constant term to the R.H.S , we have
$- 6k = - 3$.
Now we will isolate the term $k$ , it gives:
$\therefore k = \dfrac{{ - 3}}{{ - 6}} \Rightarrow k = \dfrac{1}{2}$.
We can write this also as $k = 1:2$ .

Hence $y -$ axis divide it in the ratio $1:2$.

Note: In this type of question, we have to remember the concept of section formula. We should note that if we get the negative value of $k$ , in the above solution then it suggests that the point dividing it in $k:1$ will not lie in between the line segments and it will divide the line externally, not internally.