Question
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.
$x - 3y - 2 = 0 .....(i)$
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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