Exponents relating to linear equations- help

[cmaro]

So I need to find the equation of the line passing through (1,∏) (∏₂,∏⁴) sorry, the two would only do sub script not super script but does represent squared.

So I had to find the gradient first, so that I could then sub that along with x and y into y=mx+c but I got stuck trying to find the gradient because there is pi4-pi2 / pi2-1.

How do i figure this out? i tried two things and ended up with pi - 1/pi and then pi2-3.. neither of these seem like gradients?

 

[cmaro]

never mind! i figured it out- it was m= pi2-3, and i ended up with an answer of y=pi2(x) which was correct! :)

 

[tacman]

The first thing to note is that the points given are not in standard form (x,y), but rather in (x,π^2) and (π^2,π^4). This means that the x-values are actually π^2 and the y-values are π^4.

To find the gradient, you can use the formula (y2-y1)/(x2-x1) where (x1,y1) and (x2,y2) are the coordinates of the two points given. In this case, it would be (π^4-π^2)/(π^2-1).

To simplify this, you can factor out a π^2 from both the numerator and denominator, giving you π^2(π^2-1)/(π^2-1). This simplifies to just π^2, so the gradient of the line is π^2.

Now, you can plug this gradient into the equation y=mx+c, along with one of the points given (it doesn't matter which one). For example, using the point (1,π), you would have π=π^2(1)+c. Solving for c, you get c=π-π^2.

Therefore, the equation of the line passing through the two given points is y=π^2x+(π-π^2).

I hope this helps and clarifies any confusion you had about using exponents in linear equations. Just remember to be careful with your notation and make sure you have the points in standard form before finding the gradient.