c
y= mx+ b [0.2em]
[-1em]
&Slope: m &y-intercept: bWe already know that the line's slope is m= 43. Since the given point, (0,- 2), is on the y-axis, we also know that b= - 2. Let's substitute these values into slope-intercept form.
y= 4/3x+( - 2) ⇔ y=4/3x-2
To graph this equation, we will start at the y-intercept, (0,2), and then use the line's slope to plot a second point. When we have two points, we can draw the line.
b To perform the transformation, we need to translate at least two points. Let's use the two points we found in Part A.
Since the vertical and horizontal translation mimics the rise and run of the line's slope, the points will inevitably travel along the original line. Therefore, the equation for this line is identical to the line from Part A.
y=4/3x-2
c Again, to translate the graph 5 units down, we have to perform this translation for at least two points on the line. Like in Part B, we will choose the points we identified on the original graph.
The translated line has the same slope but a y-intercept that is 5 units below that of the original line. With this information, we can write its equation.
y=4/3x-7
d All three lines have the same slope, but the line from Part C has a different y-intercept. Therefore, the lines from Part A and Part B are overlapping while the line from Part C is parallel with the first two lines.
e The slope of perpendicular lines are negative reciprocals. This means that if we multiply their slopes, the product equals -1.
m_1 * m_2 = -1
By substituting m_1= 43 into the formula, we can determine the slope of the perpendicular line.
Having found the slope of the perpendicular line, we can start writing its equation
|c|c|c|
Parallel Line & Slope & y-intercept
[-0.8em]
y= -3/4x+ b& -3/4 & b [0.8em]
To complete the equation, we have to determine the y-intercept as well. We can do that by substituting the given point, (12,7) into the equation and solving for b.
With this, we can complete the equation for the perpendicular line that passes through (12,7).
y=-3/4x+16
We can graph this line into the same system as the other lines to see that they are in fact perpendicular.
