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# On comparing the ratios $\dfrac{{{a_1}}}{{{a_2}}},\dfrac{{{b_1}}}{{{b_2}}}$ and $\dfrac{{{c_1}}}{{{c_2}}}$, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident. $6x - 3y + 10 = 0$; $2x - y + 9 = 0$ A.Intersect at a Point.B.Parallel.C.CoincidentD.Data Insufficient.

Last updated date: 16th Jun 2024
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Hint: Here, we will first denote the given linear equation in the form of a general linear equation. We will then find the ratio of the coefficients of $x$, coefficients of $y$ and constant term. Then we will compare all the ratios and find the type of line based on the comparison. Linear equations are the equations of first order which represents the equation of line.

We are given the linear equation $6x - 3y + 10 = 0$ and $2x - y + 9 = 0$.
Linear equation is of the general form $ax + by + c = 0$.
So, the given linear equation $6x - 3y + 10 = 0$ is of the form ${a_1}x + {b_1}y + {c_1} = 0$.
Now, the given linear equation $2x - y + 9 = 0$ is of the form ${a_2}x + {b_2}y + {c_2} = 0$
By comparing the coefficients of the given linear equation with the general linear equation, we get
${a_1} = 6,{b_1} = - 3,{c_1} = 10$ and ${a_2} = 2,{b_2} = - 1,{c_2} = 9$
Now, we will find the ratio of the coefficients of $x$, coefficients of $y$and constant term by substituting the values.
Ratio of the coefficients of $x$ $= \dfrac{{{a_1}}}{{{a_2}}} = \dfrac{6}{2}$
Dividing 6 by 2, we get
$\Rightarrow$ Ratio of the coefficients of $x$ $= \dfrac{{{a_1}}}{{{a_2}}} = 3$ …………………………………………………………. $\left( 1 \right)$
Ratio of the coefficients of $y$ $= \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{ - 3}}{{ - 1}}$
Dividing the terms, we get
$\Rightarrow$ Ratio of the coefficients of $y$ $= \dfrac{{{b_1}}}{{{b_2}}} = 3$ …………………………………………………………… $\left( 2 \right)$
Ratio of the coefficients of constant term $= \dfrac{{{c_1}}}{{{c_2}}} = \dfrac{{10}}{9}$ ……………………………….. $\left( 3 \right)$
Now, comparing the ratios of the coefficients of $x$, coefficients of $y$and constant term, we get $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$
If $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$, so we have no solution. Thus the lines represent that the linear equations are parallel.
Therefore, the lines representing the linear equations $6x - 3y + 10 = 0;2x - y + 9 = 0$ are parallel.
Thus option (B) is correct.

Note: We might make a mistake in comparing the ratios by considering only the ratios of the coefficients of $x$, coefficients of $y$ and leaving off the third ratio of the constant term. It is essential for us to take the ratio of the constant term into consideration to find the correct type of line. In order to find whether the lines are parallel or coincident, we need to consider the following points:
1.If $\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}}$, we have unique solution, so the lines intersect at a point.
2.If $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$, we have infinite solutions, so the lines are coincident.
3.If $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$, we have no solutions, so the lines are parallel.