please help please show work so i can do others ok so i have to find the equation of the linethat contains given point and is parallel to the given line then write a slope intercept form if possible
(8,-4) y=-1/2x+11
I also have to do the same for this type and im stuck too
(-5,-6) x axis
write given line as y = m x + b
y = (-1/2) x + 11
so m = slope = -1/2)
we need the same slope so our line is of form
y = (-1/2) x + b, but we need to find b
use given point to find b
-4 = (-1/2)8 + b
-4 = -4 + b
b = 0
so
y = (-1/2)x + 0 or -2 y = x
slope of x axis is 0 so
y = 0 x + b
-6 = 0 (-5) + b
b = -6
so
y = -6
horizontal line six units below x axis
To find the equation of a line parallel to a given line and passing through a given point, we need to use the concept of parallel lines having the same slope.
Given the point (8,-4) and the equation of the line y = -1/2x + 11, we can determine the slope of the given line. In this case, the slope is -1/2. So, the line we are looking for will also have a slope of -1/2.
Using the slope-intercept form of a linear equation, y = mx + b, where m is the slope and b is the y-intercept, we can substitute the values we have to find the y-intercept.
Substituting the point (8,-4) into the equation, we get:
-4 = (-1/2)(8) + b
Simplifying:
-4 = -4 + b
b = 0
Therefore, the equation for the line parallel to y = -1/2x + 11 and passing through the point (8,-4) is y = -1/2x.
For the second question, we are given the point (-5,-6) and the line is parallel to the x-axis. A line parallel to the x-axis has a constant y-value because the slope is zero.
So, for this case, the equation for the line would be y = -6.
Remember to always check the problem's constraints and conditions to make sure the solution satisfies all the requirements.