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Question: On comparing the ratios \[\dfrac{{{a_1}}}{{{a_2}}},\dfrac{{{b_1}}}{{{b_2}}}\] and \[\dfrac{{{c_1}}}{...

On comparing the ratios a1a2,b1b2\dfrac{{{a_1}}}{{{a_2}}},\dfrac{{{b_1}}}{{{b_2}}} and c1c2\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. 6x3y+10=06x - 3y + 10 = 0; 2xy+9=02x - y + 9 = 0
A.Intersect at a Point.
B.Parallel.
C.Coincident
D.Data Insufficient.

Explanation

Solution

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 xx, coefficients of yy 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.

Complete step-by-step answer:
We are given the linear equation 6x3y+10=06x - 3y + 10 = 0 and 2xy+9=02x - y + 9 = 0.
Linear equation is of the general form ax+by+c=0ax + by + c = 0.
So, the given linear equation 6x3y+10=06x - 3y + 10 = 0 is of the form a1x+b1y+c1=0{a_1}x + {b_1}y + {c_1} = 0.
Now, the given linear equation 2xy+9=02x - y + 9 = 0 is of the form a2x+b2y+c2=0{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
a1=6,b1=3,c1=10{a_1} = 6,{b_1} = - 3,{c_1} = 10 and a2=2,b2=1,c2=9{a_2} = 2,{b_2} = - 1,{c_2} = 9
Now, we will find the ratio of the coefficients of xx, coefficients of yyand constant term by substituting the values.
Ratio of the coefficients of xx =a1a2=62 = \dfrac{{{a_1}}}{{{a_2}}} = \dfrac{6}{2}
Dividing 6 by 2, we get
\Rightarrow Ratio of the coefficients of xx =a1a2=3 = \dfrac{{{a_1}}}{{{a_2}}} = 3 …………………………………………………………. (1)\left( 1 \right)
Ratio of the coefficients of yy =b1b2=31 = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{ - 3}}{{ - 1}}
Dividing the terms, we get
\Rightarrow Ratio of the coefficients of yy =b1b2=3 = \dfrac{{{b_1}}}{{{b_2}}} = 3 …………………………………………………………… (2)\left( 2 \right)
Ratio of the coefficients of constant term =c1c2=109 = \dfrac{{{c_1}}}{{{c_2}}} = \dfrac{{10}}{9} ……………………………….. (3)\left( 3 \right)
Now, comparing the ratios of the coefficients of xx, coefficients of yyand constant term, we get a1a2=b1b2c1c2\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}
If a1a2=b1b2c1c2\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 6x3y+10=0;2xy+9=06x - 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 xx, coefficients of yy 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 a1a2b1b2\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}}, we have unique solution, so the lines intersect at a point.
2.If a1a2=b1b2=c1c2\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 a1a2=b1b2c1c2\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.