L3.1: General equation of line

[Ankit Singh]

Hello,

I am confused how we have derived General forms for different forms of equation of line. I do not come from a great mathematics background and would really appreciate if someone would explain the concept rather than running through it.

Say for (y-y0)=m(x-x0) 

Can also be written as:

-mx+y-y0+mx0=0

Now, coeffs of above line, in the general form ax+bx+c should be:

a=-mx, b=1, c=-y0+mx0

Similarly, for a line y=mx+c and y=m(x-d) it should be:

a=-m and c=-c for y intercept

a=-m and c=md for x intercept

However in the video session this isn't explained very well and just runs off to A/B and C/B and all...

Please help me understand this and thanks!

[subhajit POD]

The general form of eq. of st. line is ax+by+c=0 

Now you have seen two point from in week 2 lectures,  (y-y0)=m(x-x0)

Hence we can write it as, mx-y+(y0-mx0)=0 

So if we compare this with the general form we get, 

a= m (i.e. the coefficient of x) 

b= -1 (i.e the coefficient of y)

c=y0-mx0 ( the constant term)

Now another form we came across in week 2 lectures was, y=mx+c 

We can write as, mx-y+c=0 

By comparing with the general form we get, 

a= m 

b=-1

c=c 

If you transform two point form in to slope intercept form you will get , 

y-y0= m(x-x0) 

implies, y=mx +(y0-mx0) 

Hence c is nothing but (y0-mx0) , which is nothing but the y-intercept. 

Hope it helps. 

Best

Subhajit

[Ankit Singh]

Hi Subhajit,

Thanks for your efforts in replying to my query, however I am interested to know how we reached the general form (i.e.. m=-A/B and y0-mx0= -C/B) and similarly for a=-C/A and b=-C/B for intercept form.

As while solving questions we would have to generalize the co-efficient values from the equation into Ax+Bx+C=0 form.

Similar to how professor has used the coefficients from the equation 3x-4y+12=0 to find a and b at this point in video. (I have marked the exact time-frame for viewing)

https://youtu.be/_3Iidm8NnbM?t=924

I am not sure if I am asking the wrong question here as my concept is not yet clear which I am trying to understand by re-watching the video multiple times, but I would appreciate any help that will bring clarity on this.

Thanks again for your efforts!

[subhajit POD]

Hi Ankit, Observe that the equation Ax+By+C = 0 always represents a st. line. If you put x=0 you will get y =-C/B ( where B not equals to 0 ) , so it is the y-intercept. Similarly if you put y=0 then you will get x=-C/A ( where A not equals 0 ) , so it is the x-intercept. 

You can write the equation in the form: y= (-A/B)x+ (-C/B)

Observe that it is in the form y=mx+c , so slope m = (-A/B) and y intercept is (-C/B). So it gives you the slope intercept form. 

Now if we consider the point slope form (P.S. in previous mail I have mentioned it as two point form wrongly, sorry for that), then (y-y0)=m(x-x0) can be expressed as, 

y=mx0 +(y0-mx0) 

So the slope is m and y-intercept is (y0-mx0). Hence m=(-A/B) and y-intercept is (y0-mx0)=(-C/B) 

Hope it helps. Let me know if there are any further doubts. 

Best

Subhajit