Question 409457
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Way off is putting it mildly.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ mx\ +\ b]


is the slope-intercept form, where *[tex \Large m] is the slope and *[tex \Large b] is the *[tex \Large y]-coordinate of the *[tex \Large y]-intercept -- that is, the point *[tex \Large (0,b)] where the graph intersects the *[tex \Large y]-axis.


But here the given point is *[tex \Large (-2,5)] which has a non-zero *[tex \Large x]-coordinate and is therefore certainly NOT a point on the *[tex \Large y]-axis.  Hence, *[tex \Large (-2,5)] is NOT the *[tex \Large y]-intercept, and *[tex \Large b\ \neq\ 5].


What you are trying to use is a formula that is inappropriate to the situation.  Furthermore, I think that using any formula at all is inappropriate to what this question is asking you to do.


Step 1:  Plot the point *[tex \Large (-2,5)].


Step 2:  Recognize that a slope of -1 means that whenever the *[tex \Large x] value increases by some amount, the *[tex \Large y] value decreases by that same amount.  You can also go the other way and increase *[tex \Large y] when decreasing *[tex \Large x], if such pleases you.


Step 3:  Starting at the point plotted in step 1, move to the right some number of units, stop, and then move down that same number of units.  (Alternatively, move to the left some number of units and then up that same number of units) Plot the new point.


Step 4:  Draw a line that passes through the two plotted points.  Done.


Now if you want the equation of the line you just graphed, that is another story and definately NOT what the question asked you to do.  But here is the process:


Since you are given an arbitrary point that is not the *[tex \Large y]-intercept and the slope, use the point-slope form of an equation of a line:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ y_1\ =\ m(x\ -\ x_1) ]


where *[tex \Large \left(x_1,y_1\right)] are the coordinates of the given point and *[tex \Large m] is the given slope.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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