Question

Prove that the following set of three lines are concurrent:
$15x-18y+1=0$; $12x+10y-3=0$ and $6x+66y-11=0$.

Answer 1

Hint: We will look at the definition of concurrent lines. We will find the point of intersection of two lines. We will find this point of intersection by solving two linear equations simultaneously. Then, we will check if this point of intersection lies on the third line. To check this, we will see whether the coordinates of the point of intersection satisfies the equation of the third line.

Complete step-by-step answer:
Concurrent lines are the lines that intersect each other at a single point. Let $15x-18y+1=0$ be equation $(i)$, $12x+10y-3=0$ be equation $(ii)$ and $6x+66y-11=0$ be equation $(iii)$. Let us solve the first equations $(i)$ and $(ii)$ by using the method of substitution. From equation $(i)$, we can write $x=\dfrac{18y-1}{15}$. Substituting this value in equation $(ii)$, we get the following,
$12\left( \dfrac{18y-1}{15} \right)+10y-3=0$
Simplifying the above equation, we get
\[\begin{align}
  & 4\left( 18y-1 \right)+50y-15=0 \\
 & \Rightarrow 72y-4+50y-15=0 \\
 & \Rightarrow 122y-19=0 \\
 & \therefore y=\dfrac{19}{122} \\
\end{align}\]
Substituting this value in equation $(i)$, we get
$15x-18\left( \dfrac{19}{122} \right)+1=0$
Simplifying this equation, we get
$\begin{align}
  & 15x-9\left( \dfrac{19}{61} \right)+1=0 \\
 & \Rightarrow 915x-171+61=0 \\
 & \Rightarrow 915x-110=0 \\
 & \Rightarrow x=\dfrac{110}{915} \\
 & \therefore x=\dfrac{22}{183} \\
\end{align}$
The coordinates of the point of intersection of lines represented by equations $(i)$ and $(ii)$ are $\left( \dfrac{22}{183},\dfrac{19}{122} \right)$. Now, we have to check if the line represented by the third equation passes through this point. We will substitute the coordinates of the point of intersection in equation $(iii)$ and see if it satisfies the equation, in the following manner,
$\begin{align}
  & \text{LHS = }6x+66y-11 \\
 & =6\times \dfrac{22}{183}+66\times \dfrac{19}{122}-11 \\
 & =\dfrac{44}{61}+\dfrac{627}{61}-11 \\
 & =\dfrac{671}{61}-11 \\
 & =\dfrac{671-671}{61} \\
 & =0 \\
 & =\text{ RHS}
\end{align}$
This shows that the point of intersection for all three lines is the same point, $\left( \dfrac{22}{183},\dfrac{19}{122} \right)$.
Hence, the given three lines are concurrent. The below figure makes it more clear.

Note: There are other ways to solve the first two linear equations. We used the substitution method. The other methods are graphing and Gauss elimination methods. We should be careful while doing the calculations so that we can avoid making any errors. It is useful to do the calculations explicitly in such types of questions.