Answer:
An equation of a line that passes through the point (4,3) and is perpendicular to the graph of the equation will be:
- [tex]y=\frac{1}{13}x+\frac{35}{13}[/tex]
Step-by-step explanation:
We know that the slope-intercept form of the line equation is
[tex]y=mx+b[/tex]
where
Given the line
y = -13x+4
comparing with the slope-intercept form of the line equation
The slope of the line = m = -13
We know that a line perpendicular to another line contains a slope that is the negative reciprocal of the slope of the other line, such as:
slope = m = -13
Thus, the slope of the the new perpendicular line = – 1/m = -1/-13 = 1/13
Using the point-slope form of the line equation
[tex]y-y_1=m\left(x-x_1\right)[/tex]
where
- m is the slope of the line
substituting the values of the slope m = 1/13 and the point (4, 3)
[tex]y-y_1=m\left(x-x_1\right)[/tex]
[tex]y-3=\frac{1}{13}\left(x-4\right)[/tex]
Add 3 to both sides
[tex]y-3+3=\frac{1}{13}\left(x-4\right)+3[/tex]
[tex]y=\frac{1}{13}x-\frac{4}{13}+3[/tex]
[tex]y=\frac{1}{13}x+\frac{35}{13}[/tex]
Therefore, an equation of a line that passes through the point (4,3) and is perpendicular to the graph of the equation will be:
- [tex]y=\frac{1}{13}x+\frac{35}{13}[/tex]
