The points (-2,0) and (0,-6) are each on the graph of a linear equation. Is (2,6) alsoon the graph of this linear equation? Explain your reasoning.

Answer:No, it does not. See below.Step-by-step explanation:Lets find out the linear equation that passes trough (-2, 0) and (0, -6). We know every linear equation has the form: y = mx + b Where m is the slope on the curve and b the independent term. We know that, given 2 points (x1,y1) and (x2,y2) we can find the slope m as:m = (y2-y1)/(x2-x1) In our case lets replace (x1,y1) and (x2,y2) by (-2, 0) and (0, -6) (notice it could be done in the inverse sense where (-2, 0) is (x2,y2) and (0, -6) is (x1,y1) ). So, our slope is:m = [-6 - 0] / [0 - (-2)] m = -6/2 = -3So, we have a downward linear function with slope -3, this is:y =  -3x + bNow, for finding b just replace any of the 2 points given in the equation. Lets replace (-2, 0):0 = -3(-2) + b0 = 6 + bSubtracting 6 in both sides:-6 = bSo, our independent term is -6 and the function is:y = -3x - 6Now lets see if this linear eqution passes trough (2,6). If it does, we can replace the values on the equation. Replacing x by 2:y = -3(2) - 6 = = -6 - 6 = - 12So, in our equation, we x is 2 y is -12, and not 6 as in the point (2,6). So, our equation does not passes trough (2,6)