Question

The graphs of the equations $5x-15y=8$ and $3x-9y=\frac{24}{9}$ are two lines which are:

• A. Coincident
• B. Parallel
• C. Intersecting exactly at one point
• D. Perpendicular to each other

Answer 1

The correct option is B

Parallel

Step 1: Find the value of ${\mathit{a}}_{\mathbf{1}}\mathbf{,}\mathbf{}{\mathit{b}}_{\mathbf{1}}\mathbf{,}\mathbf{}{\mathit{c}}_{\mathbf{1}}$ and ${\mathit{a}}_{\mathbf{2}}\mathbf{,}\mathbf{}{\mathit{b}}_{\mathbf{2}}\mathbf{,}\mathbf{}{\mathit{c}}_{\mathbf{2}}$.

It is given that the equations are $5x-15y=8$ and $3x-9y=\frac{24}{9}$.

Let $a_{1}x+b_{1}y+c_1=0$ , and $a_{2}x+b_{2}y+c_2=0$.

Rewrite the equations as:

$5x-15y-8=0$

$3x-9y-\frac{24}{9}=0$

Therefore, $a_1=5,b_1=-15,c_1=-8$ , and $a_2=3,b_2=-9,c_2=-\frac{24}{9}$.

Step 2: Check for the condition.

Here,

$\begin{array}{rcl}\frac{a_1}{a_2}& =& \frac{5}{3}\\ \frac{b_1}{b_2}& =& \frac{-15}{-9}=\frac{5}{3}\\ \frac{c_1}{c_2}& =& \frac{-8}{-{\displaystyle \frac{24}{9}}}=\frac{9}{3}\end{array}$

We get, $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$.

Hence, the lines are parallel.

Hence, option (b) is correct.