Question

Find the ratio in which the line joining $( - 2,5)$ and $( - 5, - 6)$ is divided by the line $y = - 3$. Hence find the point of intersection.

Answer 1

Hint:
To find the point of intersection of lines we will first find the ratio in which the line $y = - 3$ cuts the other line by the formula:
\[ \Rightarrow y =\dfrac{{{m_1}{y_2} + {m_2}{y_1}}}{{{m_1} + {m_2}}}\] …….(1)
And then we can find the equation of line passing through the given coordinates by the intercept formula:
$ \Rightarrow y - {y_1} =\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}(x - {x_1})$
Now, by putting the value of $y = - 3$ in the equation formed from above method we can find the point of intersection.

Complete step by step solution:
We have given that the line $y = - 3$ cuts another line whose coordinates are $( - 2,5)$ and $( - 5, - 6)$.
First of all, we will find the ratio in which line $y = - 3$cuts another line of coordinates $( - 2,5)$ and $( - 5, - 6)$.
Here ${x_1} = - 2$, ${x_2} = - 5$, ${y_1} = 5$, ${y_2} = - 6$and $y = - 3$. By applying formula in equation (1) we get,
\[ \Rightarrow y =\dfrac{{{m_1}{y_2} + {m_2}{y_1}}}{{{m_1} + {m_2}}}\]
\[ \Rightarrow ( - 3) =\dfrac{{{m_1}( - 6) + {m_2}(5)}}{{{m_1} + {m_2}}}\]
By opening the brackets, we get,
\[ \Rightarrow - 3 =\dfrac{{5{m_2} - 6{m_1}}}{{{m_1} + {m_2}}}\]
Taking the denominator on R.H.S to L.H.S and it became numerator in L.H.S we get,
\[ \Rightarrow - 3\left( {{m_1} + {m_2}} \right) = 5{m_2} - 6{m_1}\]
\[ \Rightarrow - 3{m_1} - 3{m_2} = 5{m_2} - 6{m_1}\]
Taking the like terms one side and other like terms other we get,
\[\begin{gathered}
   \Rightarrow - 3{m_1} + 6{m_1} = 5{m_2} + 3{m_2} \\
   \Rightarrow 3{m_1} = 8{m_2} \\
\end{gathered} \]
By solving this we get,
$ \Rightarrow\dfrac{{{m_1}}}{{{m_2}}} = 8:3$
To find the equation of other line we will use intercept formula i.e. equation 2 we get,
$ \Rightarrow y - {y_1} =\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}(x - {x_1})$
By putting the values of all known parameters, we get,
$ \Rightarrow y - 5 =\dfrac{{ - 6 - 5}}{{ - 5 - ( - 2)}}(x - ( - 2))$
By opening the bracket, we get,
$ \Rightarrow y - 5 =\dfrac{{ - 6 - 5}}{{ - 5 + 2}}(x + 2)$
By solving numerator and denominator we get,
$ \Rightarrow y - 5 =\dfrac{{ - 11}}{{ - 3}}(x + 2)$
Cancel the negative sign and taking denominator to other side we get,
$ \Rightarrow 3(y - 5) = 11(x + 2)$
By opening the bracket, we get,
$ \Rightarrow 3y - 15 = 11x + 22$
By taking all the terms on one side we get,
$ \Rightarrow 11x - 3y + 37 = 0$
To find the point of intersection of these lines we will put $y = - 3$ we get,
$\begin{gathered}
   \Rightarrow 11x - 3( - 3) + 37 = 0 \\
   \Rightarrow 11x + 9 + 37 = 0 \\
\end{gathered} $
Taking x on one side and constant on other we get,
$\begin{gathered}
   \Rightarrow 11x = - 46 \\
   \Rightarrow x =\dfrac{{ - 46}}{{11}} \\
\end{gathered} $

Therefore, the point of intersection is $(\dfrac{{ - 43}}{{11}}, - 3)$ and line $y = - 3$ cuts the line in $8:3$ ratio.

Note:
Students can make mistakes while calculating the equation, they usually get confused in putting the values. So, take care of the coordinates taken in the equation. Secondly, while calculating the point of intersection they get confused why we put the value of $y = - 3$ in the equation when we are going to find a point of intersection? But it is given that $y = - 3$ cuts the line. So, in this we already know the value of y and hence can find x. Students make mistakes when we have given $y = - 3$, then they take $x = 0$and your answer gets wrong. take care of these things.