1. ## Finding linear function.

"Line crosses points M and N:
M(2;6), N(-2;-6), find the equation for it"
The only formula I found was $y-y1=\frac{y2-y1}{x2-x1}(x-x1)$
It works, but this is grade 9 (in our country) exercise, I think there should be a simpler way. Is there?

2. In essence, any method you use is going to compute what you have there. The only equation for a line that I ever use is y = mx + b. For the case of two points on the line, I just plug both points into the y = mx + b equation to obtain two simultaneous equations. I solve that using any of the many valid methods, and then I'm done.

3. Originally Posted by Evaldas
"Line crosses points M and N:
M(2;6), N(-2;-6), find the equation for it"
The only formula I found was $y-y1=\frac{y2-y1}{x2-x1}(x-x1)$
It works, but this is grade 9 (in our country) exercise, I think there should be a simpler way. Is there?
that is probably the easiest method ... you could find the slope, then use y = mx+b , use either of the points and solve for b.

4. i guess you can just use observation for this problem. Notice that M is (2,6) and N is (-2,-6). Easily we can see that 6 is 3 times of 2 and -6 is also 3 times of negative two. Since it is a straight line, we can deduce that the line will be y=3x.