3. Slope-Intercept Form (Part 2)

In this lesson, we’ll derive the slope-intercept form of the line, taking different orientations of the same. This is something that I discussed in the previous lesson.

Case 1

We’ll start by taking P(x, y) in the second quadrant.

straight line slope intercept form

We’ve also constructed PA parallel to the Y-axis and PB parallel to the X axis.

Now,

OB = PA = y

BC = OC – OB = c – y

Also,

PB = -x

Now, let’s try to relate the coordinates of P with the given slope and intercept. In ΔPBC, we have:

tanθ = CB/PB = (c – y)/(-x)

And, that’s all. On rearranging, we’ll get:

y = (tanθ)x + c

or

y = mx + c

We got the same equation. Nice!

Case 2

Let’s try taking P(x, y) in the third quadrant.

straight line slope intercept form

Again, we’ve constructed PA parallel to the Y-axis and PB parallel to the X axis.

In this case,

OB = PA = -y

BC = OC + OB = c + (-y)

Also,

PB = -x

Now, let’s try to relate the coordinates of P with the given slope and intercept. In ΔPBC, we have:

tanθ = CB/PB = (c – y)/(-x)

And, on rearranging, we’ll get:

y = (tanθ)x + c

or

y = mx + c

Same equation! Still not convinced? Let’s take another case.

Case 3

This time, we’ll take a line that makes an obtuse angle with the X-axis.

Once again, we’ve constructed PA parallel to the Y-axis and PB parallel to the X axis.

Now,

OB = PA = -y

CB = OC + OB = c + (-y)

Also,

PB = x

Now, let’s try to relate the coordinates of P with the given slope and intercept. In ΔPBC, we have:

tan(π – θ) = CB/PB = (c – y)/x

And, on rearranging, we’ll get:

y = (tanθ)x + c

or

y = mx + c

Same. Equation. I know you’re convinced now. Still, why don’t you try taking some more orientations of the line and different positions of P on it? You should get the same equation in each case.

I’ll end this lesson now. The next one will discuss a few examples related to the slope-intercept form. See you there!

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