2 Marks Questions

Question 12 Marks

The line represented by x = 7 is parallel to the x-axis. Justify whether the statement is true or not.

Answer

The line represented by x = 7 is of the form x = a. The graph of the equation is a line parallel to the y-axis.
Hence, the given statement is not true.

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Question 22 Marks

For the pair of equations λx + 3y = -7 and 2x + 6y = 14 to have infinitely many solutions, the value of λ should be 1. Is the statement true? Give reasons.

Answer

λx + 3y = -7 = 0 .....(i)
2x + 6y = 14 = 0 .....(ii)
For infinitely many solutions,
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\therefore\ \frac{\text{a}_1}{\text{a}_2}=\frac{\lambda}{2}$, $\frac{\text{b}_1}{\text{b}_2}=\frac{3}{6}=\frac{1}{2}$, $\frac{\text{c}_1}{\text{c}_2}=\frac{7}{-14}=\frac{1}{-2}$
So, $\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$, for any value of $\lambda$.
Hence, the given statement is not true.

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Question 32 Marks

Two straight paths are represented by the equations $x - 3y = 2$ and $-2x + 6y = 5$. Check whether the paths cross each other or not.

Answer

Given linear equations are
$x - 3y - 2 = 0 .....(i)$
and $-2x + 6y - 5 = 0 .....(ii)$
On comparing both the equations with $ax + by + c = 0$, we get
$a_1 = 1, b_1 = -3$ and $c_1 = -2$ [from Eq. (i)]
$a_2 = -2, b_2 = 6$ and $c_2 = -5$ [from Eq . (ii)]
Here, $\frac{\text{a}_1}{\text{a}_2}=\frac{1}{-2}$
$\frac{\text{b}_1}{\text{b}_2}=\frac{-3}{6}=-\frac{1}{2}$ and $\frac{\text{c}_1}{\text{c}_2}=\frac{-2}{-5}=\frac{2}{5}$
i.e., $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$ [parallel lines]
Hence, two straight paths represented by the given equations never cross each other, because they are patallel to each other.

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Question 42 Marks

Are the following pair of linear equations consistent? Justify your answer:
x + 3y = 11 and 2(2x + 6y) = 22.

Answer

For consistent system of linear equations a, $\text{b}\neq0$
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$ (infinitely many solutions)
$\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}$ (unique solurion)
x + 3y = 11 and 2(2x + 6y) = 22
or x + 3y = 11 and 4x + 12y = 22
Here, $\frac{\text{a}_1}{\text{a}_2}=\frac{1}{4}$, $\frac{\text{b}_1}{\text{b}_2}=\frac{3}{12}=\frac{1}{4}$ and $\frac{\text{c}_1}{\text{c}_2}=\frac{11}{22}=\frac{1}{2}$
$\therefore\ \frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$
So, the given pair of linear equations is inconsistent and has no solution.

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