Question

How do you name the slope and the y-intercept of \[4x-8y=16\]?

Answer 1

Hint: The standard form of the equation of the straight line is \[ax+by+c\]. We can find the slope and intercepts of the straight line using the equation. The slope of the line having equation \[ax+by+c\] is \[\dfrac{-a}{b}\]. The Y-intercept of the line is the point where the line crosses the Y-axis, which means that the X-coordinate of the point is zero.

Complete step by step answer:
We are given the equation \[4x-8y=16\], we need to find the slope and Y-intercept. First, we need to express the equation in its standard form, to do this we need to take all the terms to one side of the equation. We can do this by subtracting 16 from both sides of the given equation. By doing this, we get
\[\begin{align}
  & \Rightarrow 4x-8y-16=16-16 \\
 & \Rightarrow 4x-8y-16=0 \\
\end{align}\]
Comparing the standard form of the equation \[ax+by+c\], we get \[a=4,b=-8\And c=-16\]. We know that the slope of the line is \[\dfrac{-a}{b}\]. Substituting the values of the coefficients, we get
\[\Rightarrow \dfrac{-4}{-8}\]
Canceling out \[-4\] as a common factor of numerator and denominator, we get
\[\Rightarrow slope=\dfrac{1}{2}\]
The Y-intercept is the point at which the line crosses the Y-axis. So, the X-coordinate is zero, substituting \[x=0\] in the equation of the straight line. We get
\[\Rightarrow 4(0)-8y-16=0\]
\[\Rightarrow -8y-16=0\]
Adding 16 to both sides, we get
\[\Rightarrow -8y-16+16=0+16\]
\[\Rightarrow -8y=16\]
Dividing both sides of the equation by \[-8\], we get
\[\begin{align}
  & \Rightarrow \dfrac{-8y}{-8}=\dfrac{16}{-8} \\
 & \therefore y=-2 \\
\end{align}\]
Hence, the slope of the line is \[\dfrac{1}{2}\], and the Y-intercept of the line is \[\left( 0,-2 \right)\].
The graph of the line is as follows,

Note:
We can also use the slope-intercept form of the equation to find the slope and y-intercept of the line. The slope-intercept form of the line is \[y=mc+c\], here \[m\] is the slope of the line and \[c\] is the Y-intercept of the line.