0

Take the line: $$y=x+1$$

I noticed that by multiplying both sides by $x-a$ the resulting equation when graphed shows the two lines: $y=x+1$ and $x=a$.

$$y(x-a)=(x+1)(x-a)$$ This works nicely when one of the lines is a vertical line but what about when neither of the lines are vertical.

How would you express say $y=2x-1$ and $y=x+1$ in one equation ?

I have tried combining them in a variety of ways but have not achieved the desired result.

Blue
  • 75,673
Kantura
  • 2,721
  • 1
    There's nothing special about having lines here. An equation of the form $pq=0$ is satisfied when $p=0$ or $q=0$. Writing your first line equations as $y-x-1=0$ and $x-a=0$, those lines together become solution set of $(y-x-1)(x-a)=0$. Likewise, writing your second equations as $y-2x+1=0$ and $y-x-1=0$, they become the solution set of $(y-2x+1)(y-x-1)=0$. – Blue Jul 23 '19 at 18:12

3 Answers3

4

You can express as $(y-2x+1) (y-x-1) = 0$.

Vamshi Kumar Kurva
  • 507
4

The general equation of a line is $$ax+by+c=0.$$ Say now you want two different lines, $a_1x+b_1y+c_1=0$ as well as $a_2x+b_2y+c_2=0$. We can multiply both expressions to obtain $$ (a_1x+b_1y+c_1)(a_2x+b_2y+c_2)=0, $$ which is satisfied by all the points of the first line as well as the second one.

Hope this helps!

R_B
  • 1,589
0

In your case you are multiplying the $(x-a)$ to both side. In the same way if you multiply an equation to both side of another equation you will get two lines. like $(y-x-1)y=(2x+1)(y-x-1)$ on multiplying $y-x-1~~ (y=x+1)$ to both side or $y(y-2x-1)=(x+1)(y-2x-1)$ Plot you can see at: -here https://www.desmos.com/calculator/pfdxywf3kt

You can also look at the curve we get when we multiply equation separately (not both side) and deduce some facts such both curve will pass from point of intersection(obvious one). Here you can find this curve https://www.desmos.com/calculator/h3pboxuymb

L F
  • 3,644
Umesh Kumar Kumawat
  • 9