What is an equation of the line having the given slope and containing the given point

M = -8,(4,6)

Recall the point-slope form of an equation:

y - y1 = m(x - x1)
where
m = slope
(x1, y1) = point on the line
Substituting,
y - 6 = -8(x - 4)
y - 6 = -8x + 32
y = -8x + 38 (slope-intercept form), or
8x + y = 38 (standard form), or
8x + y - 38 = 0 (general form)

Hope this helps~ :3

To find the equation of a line with a given slope and containing a given point, you can use the point-slope form of a linear equation. The point-slope form is given by:

y - y1 = m(x - x1)

Where (x1, y1) is the given point and m is the given slope.

In this case, the slope is -8 and the point is (4, 6). Plugging these values into the point-slope form, we get:

y - 6 = -8(x - 4)

Now, let's simplify the equation:

y - 6 = -8x + 32

Finally, we can rewrite the equation in slope-intercept form, which is y = mx + b, where m is the slope and b is the y-intercept:

y = -8x + 38

Therefore, the equation of the line with a slope of -8 and containing the point (4, 6) is y = -8x + 38.