Question

Write the equation of a line parallel to x – axis and passing through the point (−3, −4) .

Answer 1

Answer:

Equation of x− axis is y=0.

Now equation of line parallel to x− axis is y=c.

Then we get, c=−4.

So the equation of the line is y=−4.

Step-by-step explanation:

i hope it will help you ☺️❤️✌️

Answer 2

We know that , slope of line = tanθ

Where θ is angle between the line and x-axis.

If line is parallel to x-axis then angle between the line and x-axis is zero.

Let slope of given line be m.

$\rm \longrightarrow \large \: m = tan \: 0 \\ \\ \rm \longrightarrow \large \: m = 0$

$\rm \: Equation \: of \: line \: passing \: through \: point \: (x_1 , y_1) \: and \: has \: slope \: \bold{b} : \\ \\ \boxed{ \rm \large\longrightarrow y - y_1 =( x-x_1)b}$

Now, we have to find equation of line passing through point (−3, −4) and has slope 0.

$\rm \large\longrightarrow y - ( - 4) = \bigg( x-( - 3) \bigg)0 \\ \\ \\ \rm \large\longrightarrow y + 4 = (x + 3)0\\ \\ \\ \rm \large\longrightarrow y + 4 =0\\ \\ \\ \rm \large\longrightarrow y = - 4$

Answer :

Required equation of line :- y+4 = 0

Additional Information :-

$\rm \: Equation \: of \: line \: passing \: through \: points \: (x_1 , y_1) \: and \: (x_2 , y_2) : \\ \\ \boxed{ \tt \longrightarrow y - y_1 =( x-x_1)\bigg( \dfrac{y_2-y_1}{ x_2-x_1} \bigg )} \\ \\ \\ \tt \rightarrow \: here \: \bigg( \dfrac{y_2-y_1}{ x_2-x_1} \bigg ) is \: slope \: of \: line$