In this lesson, I’ll be deriving the slope-intercept form of the line here as well, but using different points on the line, as well as different orientations of the line.
The purpose is to make it clear that it is sufficient to take just one point and derive the equation, and the general form of the equation will not change if we change the position of the point.
Case I: P taken in the second quadrant
The constructions remain the same, as in the previous part. The difference is that CB is now c-y (instead of y-c). Secondly, PB=-x, instead of x. Why?
Because, x is negative in the second quadrant, therefore the distance PB (which is positive) will be -x.
Now, applying the same trigonometry in triangle PBC, we have tanθ=CB/PB=(c-y)/(-x)
On rearranging the equation, we get the same equation y=mx+c (where m=tanθ). Sound good?
Case II: P taken in the third quadrant
Same constructions, yet again! But PA=-y now, as y is negative in the 3rd quadrant.
And, CB=CO+OB=c+(-y). And, tanθ=CB/PB=(c-y)/(-x). And, we get the same equation back! y=mx+c
Hmm.. maybe you’re still not convinced. Lets take another case.
Case III: Obtuse angle
Remember that the angle θ is always measured in the anticlockwise direction, as shown in Fig. 3.
Consider triangle PBC again, tan(π-θ)=CB/PB=(y-c)/(-x)… and we get the same equation y=mx+c !
Last Case!: Negative ‘c’
What if the y-intercept or ‘c’ was negative? It doesn’t matter. We’ll get the same equation.
Here OC=-c (again, as ‘c’ is negative)
Our friend, triangle PBC has the same thing to tell us once again. tanθ=CB/PB=(c-y)/(-x), and.. sigh.. we get the same equation yet again: y = mx + c
The point has been made. The only thing to notice was the consideration of signs of the x and y coordinates, when we wanted to take distances.