Question

How do you write an equation with the slope $-3$ and contains the point $\left( 0,-3 \right)?$

Answer 1

Hint: (1) A slope of line and points are given the always prefer the form.
$\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)$
(2) The ${{x}_{1}}$ and ${{y}_{1}}$ are called. $x-$intercept and $y-$intercept.
(3) $x-$intercept is the point where the line cuts the $y-$axis.

Complete step by step solution:It is given in the question that a line has a slope at $-3$ and a line is passing through the point $\left( 0,-3 \right)$
Equation of line in mathematics can be written in two formats.
(1) $y=mx+c$
Where, $m$ is slope and $c$ is constant
(2) $\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)$
Where, $m$ is slope and $\left( {{y}_{1}},{{x}_{1}} \right)$ are the points through which the line is passing.
In question we have the slope at a line $-3,$ and the point $\left( 0,-3 \right)$ through which the line is passing.
So, we have to use the format
$\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)..(i)$
Here ${{y}_{1}}=-3,{{x}_{1}}=0$
$m=-3$
Putting above values in equation $(i)$ we get,

$\left[ y-\left( -3 \right) \right]=-3\left( x-0 \right)$
$y+3=-3x$
$3x+y+3=0$
This is the required line of equation which has slope $-3$ and passing through the point $\left( 0,-3 \right)$
The general equation of a straight line is $ax+by+c=0$
Now,
Comparing above equation with $ax+by+c=0$ we get,
$a=3,b=1,c=3$

Additional Information:
(1) In question the slope $-3$ and the point $\left( 0,-3 \right)$ is given
(2) This equation of line can be solved with another method.
(3) The another method is finding the constant $c.$ For the equation $y=mx+c$ by putting $\left( 0,-3 \right)$ as $\left( x,y \right)$ and slope $m=-3$
(4) by solving with this method we will get the same equation as we got from the earlier method.

Note:
(1) for the equation $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)\left( {{x}_{1}},{{y}_{1}} \right)$ should be $\left( 0,-3 \right)$. Do not write $\left( 0,-3 \right)$ In the place of $\left( x,y \right)$
(2) Bring the final answer in the form of $ax+by+c=0$ because it is the general form of the straight line.