Question

How do you find the equation for a line with slope 5 and y intercept $ -4 $ ?

Answer 1

Hint: We first take the general equation of a line where we have the slope and y intercept form as $ y=mx+c $ . We put the given values of slope $ m=5 $ and y intercept $ -4 $ . Then we place the equation in the graph to visualise its intercept form. We can see that the graph passes through $ \left( 0,0 \right) $ .

Complete step-by-step answer:
We take the general equation of the line with the slope $ m $ as $ y=mx+c $ .
It’s given that the value of the slope for our required line is 5.
Putting the value in the equation of $ y=mx+c $ , we get $ y=5x+c $ .
As the y intercept $ -4 $ . That’s why the line passes through $ \left( 0,-4 \right) $ .
Putting the value in the equation $ y=5x+c $ , we get $ -4=5\times 0+c $ .
This gives $ c=-4 $ .
The final equation of the line becomes $ y=5x-4 $ .

Note: For this equation $ y=5x-4 $ we can convert it into the form of $ \dfrac{x}{p}+\dfrac{y}{q}=1 $ . From the form we get that the x intercept, and y intercept of the line will be p and q respectively.
The given equation is $ y=5x-4 $ . Converting into the form of $ \dfrac{x}{p}+\dfrac{y}{q}=1 $ , we get
\[\begin{align}
  & y=5x-4 \\
 & \Rightarrow 5x-y=4 \\
 & \Rightarrow \dfrac{5x}{4}+\dfrac{y}{-4}=1 \\
 & \Rightarrow \dfrac{x}{{}^{4}/{}_{5}}+\dfrac{y}{-4}=1 \\
\end{align}\]
The form is determinate which gives that the intercept value is $ -4 $ and it passes through $ \left( 0,-4 \right) $ .