How do I write one equations in slope-intercept form, and three other lines in slope-intercept form that are perpendicular to the first line ??
for ex. If I pick the equation y=2/3x+4 for the first equation. and for the second line , the slope -2/3x +4 , how about the 3rd and 4th line?
do i choose any other value of b or slope?
The slope of the 3 perpendicular lines
must be the negative RECIPRECAL of the
first line.
1. Y = (2/3)x + 4
2. Y = (-3/2)x + 4
3. Y = (-3/2)x + 5
4. Y = (-3/2)x + 6
To write an equation in slope-intercept form (y = mx + b), you need to find the slope (m) and the y-intercept (b).
For the first equation, y = (2/3)x + 4, you've already identified the slope as 2/3 and the y-intercept as 4.
To find the equations of three other lines that are perpendicular to the first line, you can follow these steps:
1. Find the negative reciprocal of the slope (m) of the first line. In this case, the slope is 2/3, so the negative reciprocal is -3/2.
2. Use the negative reciprocal as the slope (m) for the new lines.
3. Choose any value for the y-intercept (b) of these lines. It can be any real number, as long as it's different from the y-intercept of the first line.
Let's create three more equations with these guidelines:
Equation 2: y = (-3/2)x + 1 (slope = -3/2, y-intercept = 1)
Equation 3: y = (-3/2)x + 7 (slope = -3/2, y-intercept = 7)
Equation 4: y = (-3/2)x - 2 (slope = -3/2, y-intercept = -2)
These three equations have the same slope (-3/2) as the first equation but different y-intercepts (1, 7, and -2), which makes them perpendicular to the first line.