Answer:
y=-3x+5
Explanation:
Given a line L such that:
• L has y-intercept (0,5); and
,
• L is perpendicular to the line with equation y=(1/3)x+1.
We want to find the equation of the line in the slope-intercept form.
The slope-intercept form of the equation of a straight line is given as:
![\begin{equation} y=mx+b\text{ where }\begin{cases}m={slope} \\ b={y-intercept}\end{cases} \end{equation}](https://tex.z-dn.net/?f=%5Cbegin%7Bequation%7D%20y%3Dmx%2Bb%5Ctext%7B%20where%20%7D%5Cbegin%7Bcases%7Dm%3D%7Bslope%7D%20%5C%5C%20b%3D%7By-intercept%7D%5Cend%7Bcases%7D%20%5Cend%7Bequation%7D)
Comparing the given line with the form above:
![y=\frac{1}{3}x+1\implies Slope,m=\frac{1}{3}](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B1%7D%7B3%7Dx%2B1%5Cimplies%20Slope%2Cm%3D%5Cfrac%7B1%7D%7B3%7D)
Next, we find the slope of the perpendicular line L.
• Two lines are perpendicular if the product of their slopes is -1.
Let the slope of L = m1.
Since L and y=(1/3)x+1 are perpendicular, therefore:
![\begin{gathered} m_1\times\frac{1}{3}=-1 \\ \implies Slope\text{ of line L}=-3 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20m_1%5Ctimes%5Cfrac%7B1%7D%7B3%7D%3D-1%20%5C%5C%20%5Cimplies%20Slope%5Ctext%7B%20of%20line%20L%7D%3D-3%20%5Cend%7Bgathered%7D)
The y-intercept of L is at (0,5), therefore:
![y-intercept,b=5](https://tex.z-dn.net/?f=y-intercept%2Cb%3D5)
Substitute the slope, m=-3, and y-intercept, b=5 into the slope-intercept form.
![\begin{gathered} y=mx+b \\ y=-3x+5 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20y%3Dmx%2Bb%20%5C%5C%20y%3D-3x%2B5%20%5Cend%7Bgathered%7D)
The equation of line L is: