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# How do you find the slope and the intercept of $3y + 5 = 0$ ?

Last updated date: 09th Aug 2024
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Hint: The above question is based on the concept of slope intercept form. The main approach towards solving the question is to write the above given equation in the formula of straight-line equation. This can be done by shifting the terms on the other side and isolating a single term on either side and then finding the slope and the intercept.

Complete step by step solution:
One of the forms of the equation of a straight line is also called the slope intercept form. We know the equation of straight line in slope intercept form is $y = mx + c$, Where m denotes the slope of the line and c is the y-intercept of the line. The standard equation of first degree is $Ax + By + C = 0$can be written in the slope intercept form as :
$y = \left( { - \dfrac{A}{B}} \right)x - \left( {\dfrac{C}{B}} \right)$
where $m = \left( { - \dfrac{A}{B}} \right)$ and $c = \left( { - \dfrac{C}{B}} \right)$
and also $B \ne 0$.

So now the given equation is $3y + 5 = 0$. So first we need to shift the terms which are on the left-hand side towards the right hand side. Therefore, we get
$3y + 5 = 0 \\ \Rightarrow 3y = - 5 \\$
Then we need to isolate the term y so we shift the number 3 on the right hand side to the denominator.
$y = \dfrac{{ - 5}}{3}$
The slope-intercept form is $y = mx + c$.In the above equation we can write in this form as:
$\therefore y = 0 - \dfrac{5}{3}$

So here the slope becomes 0 and y-intercept becomes $\dfrac{{ - 5}}{3}$.

Note: An important thing to note is that here the $c = \dfrac{{ - 5}}{3}$ i.e., the y-intercept of the equation is $- \dfrac{5}{3}$.In the graph we plot the equation we get to know that the equation of line will cut y-axis at $- \dfrac{5}{3}$ and the slope which is 0 where the direction of the line is horizontal.