Question

Find the ratio in which the \[y\] axis divides the line segment joining the points $(5, - 6)$ and $( - 1, - 4)$. Also find the point of intersection.

Answer 1

Hint: We can find the equation of the line using the two points given. Since every point in the \[y\] axis equals zero, substituting this we get the point of intersection of this line with \[y\] axis. Now we can apply the formula for finding the point of intersection of two lines using the ratio of division. Substituting the known values, we get the ratio.

Formula used:
The equation of a line joining the points $({x_1},{y_1})$ and $({x_2},{y_2})$ is given by $\dfrac{{x - {x_1}}}{{{x_1} - {x_2}}} = \dfrac{{y - {y_1}}}{{{y_1} - {y_2}}}$.
If a point $P(x,y)$ lies on line segment joining the points $({x_1},{y_1})$ and $({x_2},{y_2})$ divides the line in the ratio $m:n$, then the point of division has the coordinates given by $P = (\dfrac{{m{x_2} + n{x_1}}}{{m + n}},\dfrac{{m{y_2} + n{y_1}}}{{m + n}})$.

Complete step-by-step answer:
We are given the points $(5, - 6)$ and $( - 1, - 4)$.
The equation of a line joining the points $({x_1},{y_1})$ and $({x_2},{y_2})$ is given by $\dfrac{{x - {x_1}}}{{{x_1} - {x_2}}} = \dfrac{{y - {y_1}}}{{{y_1} - {y_2}}}$.
So the equation of the line segment passing through these points is given by $\dfrac{{x - 5}}{{5 - ( - 1)}} = \dfrac{{y - ( - 6)}}{{ - 6 - ( - 4)}}$.
Simplifying we get, $\dfrac{{x - 5}}{6} = \dfrac{{y + 6}}{{ - 2}} \Rightarrow \dfrac{{x - 5}}{3} = \dfrac{{y + 6}}{{ - 1}}$
Cross-multiplying we get, $(x - 5) \times ( - 1) = 3(y + 6)$
$ \Rightarrow 5 - x = 3y + 18$
$ \Rightarrow x + 3y + 13 = 0$
In the\[y\] axis, every point has $x$ coordinate zero.
So we have, $3y + 13 = 0$
Subtracting $13$ from both sides we get,
$3y = - 13$
Dividing both sides by $3$ we get,
$ \Rightarrow y = \dfrac{{ - 13}}{3}$
So the point of intersection of the line segment and $y$ axis is $(0, - \dfrac{{13}}{3})$.
We are also asked to find the ratio in which the \[y\] axis divides the line segment joining the points.
Let the required ratio be $m:n$.
If a point $P(x,y)$ lies on line segment joining the points $({x_1},{y_1})$ and $({x_2},{y_2})$ divides the line in the ratio $m:n$, then the point of division has the coordinates given by $P = (\dfrac{{m{x_2} + n{x_1}}}{{m + n}},\dfrac{{m{y_2} + n{y_1}}}{{m + n}})$.
Here we have, $P = (0, - \dfrac{{13}}{3})$.
${x_1} = 5,{x_2} = - 1,{y_1} = - 6,{y_2} = - 4$
Substituting we get,
$(0, - \dfrac{{13}}{3}) = (\dfrac{{m \times - 1 + n \times 5}}{{m + n}},\dfrac{{m \times - 4 + n \times - 6}}{{m + n}})$
$(0, - \dfrac{{13}}{3}) = (\dfrac{{ - m + 5n}}{{m + n}},\dfrac{{ - 4m - 6n}}{{m + n}})$
This gives, $0 = \dfrac{{ - m + 5n}}{{m + n}}$ and $\dfrac{{13}}{3} = \dfrac{{ - 4m - 6n}}{{m + n}}$
Considering the first equation we have, $\dfrac{{ - m + 5n}}{{m + n}} = 0$
Cross-multiplying we get,
$ - m + 5n = 0$
Rearranging we get,
$m = 5n$.
This gives $\dfrac{m}{n} = \dfrac{5}{1}$.
Therefore the required ratio is $5:1$

Note: This question can be solved in an easier way. We do not need to find the equation of the line.
If a point $P(x,y)$ lies on line segment joining the points $({x_1},{y_1})$ and $({x_2},{y_2})$ divides the line in the ratio $m:n$, then the point of division has the coordinates given by $P = (\dfrac{{m{x_2} + n{x_1}}}{{m + n}},\dfrac{{m{y_2} + n{y_1}}}{{m + n}})$.
Put, $\dfrac{{m{x_2} + n{x_1}}}{{m + n}} = 0$, since the point is on the $y$ axis.
Then substituting ${x_1},{x_2}$ we get the ratio $m:n$.