Can someone explain this to me??

Question

I have the answer, but I really don't understand how you get to that point. Thanks!

A is the point (a^2, a)
T is the point (t^2, t)
a ≠ t

Find the gradient of the line AT. Give your answer in its simplest form.

It says the answer is 1/t+a. Can you explain how you get there?

Answer 1

find the gradient using these two point
m is the gradient
Y - y = m(X - x)
( t - a)= m(t^2 - a^2)
m(t^2 - a^2) = ( t - a)
m( t - a)( t + a)= ( t - a) (cancelling ( t - a) on both side)
m = 1/( t + a)

Answer 2

The gradient of the line AT is m = (a - t)/ (a^2 - t^2)

m = (a-t)/ [(a - t)(a+t)] = 1/ (a + t) so I think that your answer should read 1/(t + a) which is the same thing.

Answer 3

The slope of this line is
(t - a) / (t^2 - a^2) =
(t - a) / (t+a)(t-a) =
1 / (t+a)

So the general equation for this line is
f(x) = (1 / (t+a))x + b

Since we're only working with two dimensions, you can think of grad(f) as just . When you take the derivative of f(x) with respect to x, you just get 1/(t+a).

Answer 4

The gradient of (x1, y1) and (x2, y2) is defined as

m = (y2-y1)/(x2-x1).

By substitution, you get

m = (a - t)/ (a^2 - t^2).

Now, (a^2 - t^2) is a difference of two squares, whose factors are (a - t) and (a + t). Thus,

m = (a - t)/ [(a - t) (a + t)]

m = 1 / (a + t) since (a-t) on the numerator and denominator will cancel altogether.

Answer 5

m = (t - a) / (t² - a²)
m = (t - a) / ((t - a).(t + a))
m = 1 / (t + a)

Answer 6

A is the point (a^2, a)
T is the point (t^2, t),

Gradient
= (t-a)/(t^2 - a^2)
= (t-a)/[(t+a)(t-a)] -------note a^2-b^2 = (a+b)(a-b) always
= 1/(t+a)

Answer 7

Gradient = (Yt -Ya) /(Xt - Xa)
= (t - a) / (t^2 - a^2)
= (t - a) / ( ( t + a ) * ( t - a ))
= 1/(t+a)

It's difficult to explain here.. Hope they'll integrate Microsoft Equation 3.0 someday in Yahoo!answers

Answer 8

the slope of a line is (y1-y2)/(x1-x2)

in this problem
y1 = a and y2 = t
x1 = (a^2) and x2 = (t^2)

(a-t)/a^2-t^2

the difference of two squares (a^2 and t^2) can be rewritten as (a+t)(a-t)*.

so a-t/a^2-t^2 = (a-t)/(a+t)(a-t)

and that equals 1/(a+t).


* becuase (a+t)(a-t) = a^2 + ta - ta - t^2 = a^2 + t^2

Answer 9

c thats the right answwers becuase both of the numbers are equal to 1 so thats the correct answer