# The parallelism condition for two straight lines one of which is specified by the equation $ax+by+c=0$ the other being represented parametrically by $x=\alpha t+\beta ,y=\gamma t+\delta $ is given by

(a) $\alpha \gamma -b\alpha =0,\beta =\delta =c=0$

(b) $a\alpha -b\gamma =0,\beta =\delta =0$

(c) $a\alpha +b\gamma =0$

(d) $a\gamma =b\alpha =0$

Last updated date: 18th Mar 2023

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Answer

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Hint: Use parametric equations to find the straight line. Take the slope of both equations to form the required equation.

Given lines are $ax+by+c=0.................\left( 1 \right)$

$x=\alpha t+\beta ................\left( 2 \right)$

$y=\gamma t+\delta .............\left( 3 \right)$

One of the straight line specified is $ax+bx+c=0$

We have to find other straight line by using the equation (2) and (3)

So as to complete the parallelism condition; By cross multiplying:

$\begin{align}

& x=\alpha t+\beta \\

& y=\gamma t+\delta \\

& \dfrac{x}{y}=\dfrac{\alpha t+\beta }{\gamma t+\delta } \\

& \Rightarrow x\left( \gamma t+\delta \right)=y\left( \alpha t+\beta \right) \\

\end{align}$

Divide throughout by $t$

\[\begin{align}

& \dfrac{x\left( \gamma t+\delta \right)}{t}=\dfrac{y\left( \alpha t+\beta \right)}{t} \\

& \Rightarrow \gamma \alpha +\dfrac{\delta x}{t}=\alpha y+\dfrac{\beta y}{t} \\

\end{align}\]

$\Rightarrow \left( \gamma x-\alpha y \right)+\left( \dfrac{\delta x}{t}-\dfrac{\beta y}{t} \right)=0...................\left( 4 \right)$

Slope ${{m}_{1}}=\alpha =\dfrac{{{a}_{1}}}{{{b}_{1}}}=\dfrac{x}{t},$ for equation (1)

Slope ${{m}_{2}}=\gamma =\dfrac{{{a}_{2}}}{{{b}_{2}}}=\dfrac{y}{t}$ , for equation (2)

Substituting the slope on equation (4)

${{\gamma }_{x}}-\alpha y+\alpha \delta -\gamma \beta =0...............\left( 5 \right)$

The equation (5) is the 2nd straight line to fulfil parallelism condition.

$ax+by+c=0$

$\gamma x-\alpha y+\alpha \delta -\gamma \beta =0$ (Equation 5)

$\therefore $ The required equation is found from the slope of the equation.

From this by considering that the slope of both the equations are similar i.e., now comparing the common equation and equation (5).

$\begin{align}

& ax+by+c=0 \\

& \gamma x-\alpha y+\alpha \delta -\gamma \beta \\

& here\text{ }{{\text{a}}_{1}}=a\text{ }{{\text{a}}_{2}}=\gamma \\

& \text{ }{{\text{b}}_{1}}=a\text{ }{{\text{b}}_{2}}=\alpha \text{ } \\

\end{align}$

We can consider that

\[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\Rightarrow \dfrac{a}{\gamma }=\dfrac{b}{-\alpha }\]

Now cross multiplying them we get,

\[-a\alpha =b\gamma \Rightarrow a\alpha +b\gamma =0\]

Thus we received the required equation.

So option C is correct.

Note: The slope of a line characterizes the direction of a line. In this question it is important to find the slope to fulfil the conditions of parallelism.

To find the slope, you need to divide the difference of $y$-coordinate of two points on a line by the difference of the $x$-coordinate of those same two points.

$\therefore slope=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$

Or, slope can be found also using

$\Delta $ Notation,

$y$-coordinate $=\Delta y$

$x$-coordinate $=\Delta x$

$\therefore slope=\dfrac{\Delta y}{\Delta x}$

$\dfrac{\Delta y}{\Delta x}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}=$slope of line

Given lines are $ax+by+c=0.................\left( 1 \right)$

$x=\alpha t+\beta ................\left( 2 \right)$

$y=\gamma t+\delta .............\left( 3 \right)$

One of the straight line specified is $ax+bx+c=0$

We have to find other straight line by using the equation (2) and (3)

So as to complete the parallelism condition; By cross multiplying:

$\begin{align}

& x=\alpha t+\beta \\

& y=\gamma t+\delta \\

& \dfrac{x}{y}=\dfrac{\alpha t+\beta }{\gamma t+\delta } \\

& \Rightarrow x\left( \gamma t+\delta \right)=y\left( \alpha t+\beta \right) \\

\end{align}$

Divide throughout by $t$

\[\begin{align}

& \dfrac{x\left( \gamma t+\delta \right)}{t}=\dfrac{y\left( \alpha t+\beta \right)}{t} \\

& \Rightarrow \gamma \alpha +\dfrac{\delta x}{t}=\alpha y+\dfrac{\beta y}{t} \\

\end{align}\]

$\Rightarrow \left( \gamma x-\alpha y \right)+\left( \dfrac{\delta x}{t}-\dfrac{\beta y}{t} \right)=0...................\left( 4 \right)$

Slope ${{m}_{1}}=\alpha =\dfrac{{{a}_{1}}}{{{b}_{1}}}=\dfrac{x}{t},$ for equation (1)

Slope ${{m}_{2}}=\gamma =\dfrac{{{a}_{2}}}{{{b}_{2}}}=\dfrac{y}{t}$ , for equation (2)

Substituting the slope on equation (4)

${{\gamma }_{x}}-\alpha y+\alpha \delta -\gamma \beta =0...............\left( 5 \right)$

The equation (5) is the 2nd straight line to fulfil parallelism condition.

$ax+by+c=0$

$\gamma x-\alpha y+\alpha \delta -\gamma \beta =0$ (Equation 5)

$\therefore $ The required equation is found from the slope of the equation.

From this by considering that the slope of both the equations are similar i.e., now comparing the common equation and equation (5).

$\begin{align}

& ax+by+c=0 \\

& \gamma x-\alpha y+\alpha \delta -\gamma \beta \\

& here\text{ }{{\text{a}}_{1}}=a\text{ }{{\text{a}}_{2}}=\gamma \\

& \text{ }{{\text{b}}_{1}}=a\text{ }{{\text{b}}_{2}}=\alpha \text{ } \\

\end{align}$

We can consider that

\[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\Rightarrow \dfrac{a}{\gamma }=\dfrac{b}{-\alpha }\]

Now cross multiplying them we get,

\[-a\alpha =b\gamma \Rightarrow a\alpha +b\gamma =0\]

Thus we received the required equation.

So option C is correct.

Note: The slope of a line characterizes the direction of a line. In this question it is important to find the slope to fulfil the conditions of parallelism.

To find the slope, you need to divide the difference of $y$-coordinate of two points on a line by the difference of the $x$-coordinate of those same two points.

$\therefore slope=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$

Or, slope can be found also using

$\Delta $ Notation,

$y$-coordinate $=\Delta y$

$x$-coordinate $=\Delta x$

$\therefore slope=\dfrac{\Delta y}{\Delta x}$

$\dfrac{\Delta y}{\Delta x}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}=$slope of line

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