So far we’ve covered two forms of the equation of a straight line. Here’s another one, known as the point-slope form.
Consider a line which passes through the point (x1, y1), and has the slope m (=tanθ).
I’ll derive it’s equation using two different methods again.
We know the slope-intercept form of the line.. lets begin with that. Let the equation be y = mx + c.
Now we haven’t been given the value of c, but we know that the line passes through the point (x1, y1).
Since the relation y = mx + c holds true for all the points on the line, it must also hold true for x1 and y1.
Therefore, we can write, y1 = mx1 + c. This gives the value of c as y1 – mx1.
Substituting this back in the equation, and rearranging the terms, we get the equation as y – y1=m(x – x1)
“But I don’t know what the slope intercept form is!”. No problem. We can derive the equation from scratch, as usual. Let’s move on to the next method.
Let’s take any point P(x, y) on the line, and do some constructions as below.
Now tanθ=PQ/AQ which implies m=(y – y1)/(x – x1). (I have skipped a lot of explanation here.. go back to see how PQ and AQ are calculated)
And… we get the same equation as before y – y1=m(x – x1)
- The equation of a line having a given slope m, and passing through a given point (x1, y1), has the equation: y – y1=m(x – x1)
This form will be one of the most frequently used forms, so keep it handy. See you in the next lesson with some examples !