Slope-Intercept Form (Part 1)


Welcome! In this lesson, I’ll derive the equation to a slanted line, which looks like this:

Coordinate Geometry Slanted Line

Slope Intercept Form

Given that, the line makes a certain angle (θ) with the X axis, and passes through a point on the Y axis which is at a certain distance ‘c’ from the origin.

But do we really need both the angle and ‘c’? What if only the angle were given? What if only ‘c’ was given?

If only one of these were given, then there wouldn’t have been a unique line possible. Look at the figures below.

Coordinate Geometry Slanted Line

 

Coordinate Geometry Slanted Line

 

So we’ll need both the angle and ‘c’, for there to be a unique line. In general, as we’ll see later, we’ll require two given conditions to uniquely determine a line on the XY plane.

By the way this ‘c’ is the y-intercept, which will be positive or negative depending upon whether the point (at which the line intersects the Y axis) lies above or below the origin. (It can also be zero, of course)

And the tangent of the angle made by the line (i.e. tanθ) is known as the slope of the line. The convention is to measure the angle in the anticlockwise direction with the X-axis. (which will make the angle lie in the range [0, π) )

Now, we have a unique line, whose slope and y-intercept is given. On to its equation !

Lets take general point P(x, y) on the line, and try to find a relation which will always hold true (the definition!)

Coordinate Geometry Straight line Slope intercept form derivation

Too much of a mess there.. PA perpendicular to the X axis.. and CB parallel.. don’t run off yet !

Well? PB always equals y-c, CB always equals x, and PB/CA always equals tanθ (as PCB is a right angled triangle)

So, we have tanθ=(y-c)/x and.. that’s it ! There’s our equation !

I’ll rearrange the equation so it looks a little nice. y = mx + c (where m=tanθ)

Lesson Summary

  1. The equation of a line which makes an angle of θ (measured anticlockwise) with the X axis and has the y-intercept ‘c’ is given by y = mx + c, where m = tanθ

Wait a minute.. didn’t I just fool you by taking the point P of my choice? and an angle of my choice? and ‘c’ to be positive?

What if the angle was obtuse? What if P was taken somewhere on the left of the Y axis? And if ‘c’ was negative?

Fortunately, the equation will remain the same, as the method remains the same. I’ll explain this in the next lesson, by taking different orientations of the line, values of ‘c’, and positions of P.


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