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Q6(ii):
Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is: (ii) parallel lines

Solution :

Given: A linear equation in two variables, $2x + 3y - 8 = 0$.

To Find: Another linear equation in two variables such that the pair of equations represents parallel lines.

Theoretical Background:

For a pair of linear equations in two variables of the form:

$a_1x + b_1y + c_1 = 0$

$a_2x + b_2y + c_2 = 0$

The lines are parallel if and only if the ratios of the coefficients of $x$ and $y$ are equal, but the ratio of the constant terms is different. Mathematically, the condition is:

$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$

Step 1: Identify the coefficients of the given equation.

Given equation: $2x + 3y - 8 = 0$

Here, $a_1 = 2$, $b_1 = 3$, and $c_1 = -8$.

Step 2: Determine the coefficients for the second equation.

To satisfy the condition $\frac{a_1}{a_2} = \frac{b_1}{b_2}$, we can multiply $a_1$ and $b_1$ by a common constant $k$. Let us choose $k = 2$.

$a_2 = 2 \times 2 = 4$

$b_2 = 3 \times 2 = 6$

Now, we must ensure $\frac{c_1}{c_2} \neq \frac{a_1}{a_2}$. Since $\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$, we must choose $c_2$ such that $\frac{-8}{c_2} \neq \frac{1}{2}$.

If we choose $c_2 = -9$, then $\frac{-8}{-9} = \frac{8}{9}$, which is clearly not equal to $\frac{1}{2}$.

Step 3: Formulate the second equation.

Using the coefficients $a_2 = 4$, $b_2 = 6$, and $c_2 = -9$, the equation is:

$4x + 6y - 9 = 0$

Verification:

$\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$

$\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}$

$\frac{c_1}{c_2} = \frac{-8}{-9} = \frac{8}{9}$

Since $\frac{1}{2} = \frac{1}{2} \neq \frac{8}{9}$, the condition for parallel lines is satisfied.

Final Answer: One such linear equation is 4x + 6y - 9 = 0. (Note: Many such equations are possible by choosing different values for k and c2).


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