Find the ratio in which the y-axis divides the line segment joining the points (-4, -6) and (10, 12). Also, find the coordinates of the point of division.

Hint: We will first assume the ratio. Then apply the section formula and put the x-coordinate to be 0 because the point lies on the y-axis. Thus, we will get the value of the assumed ratio.

Complete step-by-step answer:
We have the points (-4, -6) and (10, 12) with us which are divided into some ratio by the y-axis.
The line formed by these points is divided by the y-axis.
Let the ratio it is divided by the y-axis is k : 1.
Let us now see the section formula which we will need to use in this solution.
If we have two points $({x_1},{y_1})$ and $({x_2},{y_2})$ which are divided in m : 1, and the point which divides it in m : 1 is $(x,y)$. Then $(x,y) = \left( {\dfrac{{m.{x_2} + {x_1}}}{{m + 1}},\dfrac{{m.{y_2} + {y_1}}}{{m + 1}}} \right)$.
Applying this to (-4, -6) and (10, 12) with k : 1.
We have $(x,y) = \left( {\dfrac{{k \times 10 - 4}}{{k + 1}},\dfrac{{k \times 12 - 6}}{{k + 1}}} \right)$.
$ \Rightarrow (x,y) = \left( {\dfrac{{10k - 4}}{{k + 1}},\dfrac{{12k - 6}}{{k + 1}}} \right)$.
Since this point lies on the y-axis.
Hence, its x-coordinate is 0.
So, $\dfrac{{10k - 4}}{{k + 1}} = 0$
This can be written as:-
$ \Rightarrow 10k - 4 = 0$
Taking the 4 from LHS to RHS, we will have:-
$ \Rightarrow 10k = 4$
Taking the 10 on LHS to RHS, we will have:-
$ \Rightarrow k = \dfrac{4}{{10}}$
Simplifying the RHS further to get the following result:-
$ \Rightarrow k = \dfrac{2}{5}$.
Hence, the required ratio is 2 : 5.
We also need to find the coordinates. So, putting the value of k in $ \Rightarrow (x,y) = \left( {\dfrac{{10k - 4}}{{k + 1}},\dfrac{{12k - 6}}{{k + 1}}} \right)$.
So, $x = \dfrac{{10 \times \dfrac{3}{5} - 4}}{{\dfrac{3}{5} + 1}} = \dfrac{{6 - 4}}{{\dfrac{{3 + 5}}{5}}} = \dfrac{2}{8} \times 5 = \dfrac{5}{4}$.
Hence, the required point is $\left( {\dfrac{5}{4},0} \right)$.

Note: The students must note that we have taken the ratio k : 1 instead of m : n because two unknown in the ratio but only one condition to solve it will not at all be helpful to us directly but it will somehow give us the ratio of m and n which will be the required ratio.
The student must note that if the question is given that the point lies on the x –axis, then we will take the y-coordinate to be equal to 0 and find the required answer.