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Question: Find the ratio in which the \[y\] axis divides the line segment joining the points \((5, - 6)\) and ...

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

Explanation

Solution

We can find the equation of the line using the two points given. Since every point in the yy axis equals zero, substituting this we get the point of intersection of this line with yy 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 (x1,y1)({x_1},{y_1}) and (x2,y2)({x_2},{y_2}) is given by xx1x1x2=yy1y1y2\dfrac{{x - {x_1}}}{{{x_1} - {x_2}}} = \dfrac{{y - {y_1}}}{{{y_1} - {y_2}}}.
If a point P(x,y)P(x,y) lies on line segment joining the points (x1,y1)({x_1},{y_1}) and (x2,y2)({x_2},{y_2}) divides the line in the ratio m:nm:n, then the point of division has the coordinates given by P=(mx2+nx1m+n,my2+ny1m+n)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)(5, - 6) and (1,4)( - 1, - 4).
The equation of a line joining the points (x1,y1)({x_1},{y_1}) and (x2,y2)({x_2},{y_2}) is given by xx1x1x2=yy1y1y2\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 x55(1)=y(6)6(4)\dfrac{{x - 5}}{{5 - ( - 1)}} = \dfrac{{y - ( - 6)}}{{ - 6 - ( - 4)}}.
Simplifying we get, x56=y+62x53=y+61\dfrac{{x - 5}}{6} = \dfrac{{y + 6}}{{ - 2}} \Rightarrow \dfrac{{x - 5}}{3} = \dfrac{{y + 6}}{{ - 1}}
Cross-multiplying we get, (x5)×(1)=3(y+6)(x - 5) \times ( - 1) = 3(y + 6)
5x=3y+18\Rightarrow 5 - x = 3y + 18
x+3y+13=0\Rightarrow x + 3y + 13 = 0
In theyy axis, every point has xx coordinate zero.
So we have, 3y+13=03y + 13 = 0
Subtracting 1313 from both sides we get,
3y=133y = - 13
Dividing both sides by 33 we get,
y=133\Rightarrow y = \dfrac{{ - 13}}{3}
So the point of intersection of the line segment and yy axis is (0,133)(0, - \dfrac{{13}}{3}).
We are also asked to find the ratio in which the yy axis divides the line segment joining the points.
Let the required ratio be m:nm:n.
If a point P(x,y)P(x,y) lies on line segment joining the points (x1,y1)({x_1},{y_1}) and (x2,y2)({x_2},{y_2}) divides the line in the ratio m:nm:n, then the point of division has the coordinates given by P=(mx2+nx1m+n,my2+ny1m+n)P = (\dfrac{{m{x_2} + n{x_1}}}{{m + n}},\dfrac{{m{y_2} + n{y_1}}}{{m + n}}).
Here we have, P=(0,133)P = (0, - \dfrac{{13}}{3}).
x1=5,x2=1,y1=6,y2=4{x_1} = 5,{x_2} = - 1,{y_1} = - 6,{y_2} = - 4
Substituting we get,
(0,133)=(m×1+n×5m+n,m×4+n×6m+n)(0, - \dfrac{{13}}{3}) = (\dfrac{{m \times - 1 + n \times 5}}{{m + n}},\dfrac{{m \times - 4 + n \times - 6}}{{m + n}})
(0,133)=(m+5nm+n,4m6nm+n)(0, - \dfrac{{13}}{3}) = (\dfrac{{ - m + 5n}}{{m + n}},\dfrac{{ - 4m - 6n}}{{m + n}})
This gives, 0=m+5nm+n0 = \dfrac{{ - m + 5n}}{{m + n}} and 133=4m6nm+n\dfrac{{13}}{3} = \dfrac{{ - 4m - 6n}}{{m + n}}
Considering the first equation we have, m+5nm+n=0\dfrac{{ - m + 5n}}{{m + n}} = 0
Cross-multiplying we get,
m+5n=0- m + 5n = 0
Rearranging we get,
m=5nm = 5n.
This gives mn=51\dfrac{m}{n} = \dfrac{5}{1}.
Therefore the required ratio is 5:15:1

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)P(x,y) lies on line segment joining the points (x1,y1)({x_1},{y_1}) and (x2,y2)({x_2},{y_2}) divides the line in the ratio m:nm:n, then the point of division has the coordinates given by P=(mx2+nx1m+n,my2+ny1m+n)P = (\dfrac{{m{x_2} + n{x_1}}}{{m + n}},\dfrac{{m{y_2} + n{y_1}}}{{m + n}}).
Put, mx2+nx1m+n=0\dfrac{{m{x_2} + n{x_1}}}{{m + n}} = 0, since the point is on the yy axis.
Then substituting x1,x2{x_1},{x_2} we get the ratio m:nm:n.