the equation of a line parallel to the given line but passing through the given point. I can't figure out.
y=-3x-6;(-1, 5)
y-y1 =m(x-x1)
y=5=-3x-*(x-1)
y-5=-3x-3
y-5+3=-2+1=
y=-3
I don't know if this is correct
The given line has slope -3
So, the parallel line also has that slope. Now you have a point and a line, so the equation is
y-5 = -3(x+1)
y = -3x+2
Your first line was okay, but then you started dropping x's along the way.
x1 = -1, not 1.
Thank You Steve I will note that thanks again !
To find the equation of a line parallel to a given line but passing through a specific point, you can use the fact that parallel lines have the same slope.
In this case, the given line equation is y = -3x - 6. The slope of this line is -3, which is also the slope of any line parallel to it.
Now, we need to find the equation of a line passing through the point (-1, 5) with a slope of -3.
To do this, we can use the point-slope form of a linear equation, which is given by:
y - y1 = m(x - x1)
Where (x1, y1) is the given point and m is the slope.
Substituting the values into the equation, we get:
y - 5 = -3(x - (-1))
Simplifying further:
y - 5 = -3(x + 1)
Distribute -3:
y - 5 = -3x - 3
Now, let's rearrange the equation to convert it into the slope-intercept form, y = mx + b, where m is the slope and b is the y-intercept:
y = -3x - 3 + 5
Combine like terms:
y = -3x + 2
So, the equation of the line parallel to y = -3x - 6 and passing through the point (-1, 5) is y = -3x + 2.