Question

Which of the following pairs of linear equations intersect at a point, parallel or coincident
1. $5x-4y+8=0;7x+6y-9=0$
2. $9x+3y+12=0;18x+6y+24=0$
3. $6x-3y+10=0;2x-y+9=0$

Answer 1

Hint : We first state the conditions for the given pairs of linear equations to intersect at a point, to be parallel or coincident. The three conditions to be $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}$, $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}$, $\dfrac{{{a}_{1}}}{{{a}_{2}}}\ne \dfrac{{{b}_{1}}}{{{b}_{2}}}$. We find the values for the given equations and find the solution.

Complete step by step solution:
We take two arbitrary linear equations ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0;{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0$ and the ratio of their respective coefficients. We get $\dfrac{{{a}_{1}}}{{{a}_{2}}},\dfrac{{{b}_{1}}}{{{b}_{2}}},\dfrac{{{c}_{1}}}{{{c}_{2}}}$.
Now if $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}$ satisfies, we can say the lines are coincident.
If $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}$ satisfies, we can say the lines are parallel.
If $\dfrac{{{a}_{1}}}{{{a}_{2}}}\ne \dfrac{{{b}_{1}}}{{{b}_{2}}}$ satisfies, we can say the lines intersect at a point.

We now check the condition for the given pairs of linear equations.
For $5x-4y+8=0;7x+6y-9=0$, we get the coefficients as $\dfrac{5}{7},\dfrac{-4}{6},\dfrac{8}{-9}$.
The relation for the lines is $\dfrac{{{a}_{1}}}{{{a}_{2}}}\ne \dfrac{{{b}_{1}}}{{{b}_{2}}}$. Therefore, the lines intersect at a point.

For $9x+3y+12=0;18x+6y+24=0$, we get the coefficients as $\dfrac{9}{18},\dfrac{3}{6},\dfrac{12}{24}$.
The simplified forms are $\dfrac{9}{18}=\dfrac{1}{2},\dfrac{3}{6}=\dfrac{1}{2},\dfrac{12}{24}=\dfrac{1}{2}$.
The relation for the lines is $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}$. Therefore, the lines are coincident.

For $6x-3y+10=0;2x-y+9=0$, we get the coefficients as $\dfrac{6}{2}=3,\dfrac{-3}{-1}=3,\dfrac{10}{9}$.
The relation for the lines is $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}$. Therefore, the lines are parallel.

Note : We need to first care about the coefficients of the variables. As they decide where the lines intersect or not. The condition of parallel and coincident is almost similar. The ratio of the constant differentiates them into two parts.