Write the equation of (-6, -4), (0. -4) and (6, -4)
I don't understand what you are wanting. Is this supposed to be three points which define a plane? (there are no z coordinates). Is this supposed to be three points on a conic? What type conic?
I assume the question was:
"Write the equation of the line passing through the following points,(-6, -4), (0, -4) and (6, -4)."
The three points are collinear and represent a horizontal line (because the y-value does not vary) situated at y=-4 no matter what the x-value is.
The line is therefore simply:
y=-4.
To find the equation of a line, we can use the slope-intercept form, which is given by y = mx + b, where "m" represents the slope of the line, and "b" represents the y-intercept.
First, let's find the slope (m) using the formula:
m = (y₂ - y₁) / (x₂ - x₁)
Using the points (-6, -4) and (0, -4):
m = (-4 - (-4)) / (0 - (-6))
m = 0 / 6
m = 0
Now that we have the slope, we need to find the y-intercept (b). We can use any of the given points to do this. Let's use the point (0, -4):
y = mx + b
-4 = 0 * 0 + b
-4 = b
Therefore, b = -4.
Now we have the slope (m = 0) and the y-intercept (b = -4), so we can write the equation in slope-intercept form:
y = mx + b
y = 0x - 4
y = -4
Thus, the equation of the line passing through the points (-6, -4), (0, -4), and (6, -4) is y = -4.