a) Determine an equation for the line with slope 3 and passing through the origin.
b) What is an equation for the line perpendicular to the line in part a) that also passes through the point (0,2)?
a. Given: m = 3, P(0,0).
Y = mx + b.
0 = 3*0 + b
b = 0.
Eq: Y = 3x.
b. m = -1/3, P(0,2).
Y = mx + b.
2 = (-1/3)*0 + b
b = 2.
Eq: Y = (-1/3)x + 2.
a) To determine the equation for the line with slope 3 and passing through the origin, we can use the slope-intercept form of a linear equation: y = mx + b, where m is the slope and b is the y-intercept.
In this case, the slope (m) is given as 3 and the line passes through the origin, which means the y-intercept is 0.
So, substituting the values into the slope-intercept form, we have: y = 3x + 0
Simplifying the equation, we get: y = 3x
Therefore, the equation for the line with slope 3 and passing through the origin is y = 3x.
b) To find the equation for the line perpendicular to the line in part a) and passing through the point (0,2), we need to determine the slope of the new line.
The slope of a line perpendicular to another line is the negative reciprocal of the slope of the original line.
In part a), the original line has a slope of 3. So, the slope of the new line would be -1/3.
Now, we can use the point-slope form of a linear equation: y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line.
Substituting the values into the point-slope form, we have: y - 2 = (-1/3)(x - 0)
Simplifying the equation, we get: y - 2 = (-1/3)x
To get the equation in the slope-intercept form, we can further simplify: y = (-1/3)x + 2
Therefore, the equation for the line perpendicular to the line in part a) and passing through the point (0,2) is y = (-1/3)x + 2.