Mathematics, 03.08.2019 12:10 NotYourStudent

Here is the question again, for your .

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Answer from: camihecma1057

This is the one I referred to as Problem 3.

Step-by-step explanation:

The length of OP is going to require a bit more work since we don't know P yet.

We need to find the equation for the line that travels through A & B.

Then we need to find the line that travels through O & P such that the choice in P makes OP perpendicular to AB. Perpendicular lines have opposite reciprocal slopes. The cool thing about OP is it is easy to identity the y-intercept. So the line for OP will just be $y=\text{opposite reciprocal slope of }AB \cdot x$ . Slope-intercept form is y=mx+b where m is the slope and b is the y-intercept.

Let's go ahead and find the slope of AB.

The slope of AB can be found by using the formula:

$\frac{y_2-y_1}{x_2-x_1}$ where (x_1,y_1,) and (x_2,y_2) are points given to you on the line.

The points given to us are: (-1,1) and $(\frac{1}{a},\frac{1}{a^2})$

So entering these points into the formula gives us:

$\frac{\frac{1}{a ^2}-1}{\frac{1}{a}-(-1)}$

Simplifying:

$\frac{\frac{1}{a^2}-1}{\frac{1}{a}+1}$

Clearing the mini-fractions by multiplying top and bottom by a^2 :

$\frac{1-a 2}{a+a^2}$ .

The top is a difference of squares and so can use formula a^2-b^2=(a-b)(a+b) .

The bottom both terms have a common factor of so I can just factor that out of the two terms.

Let's do that:

$\frac{(1-a)(1+a)}{a(1+a)}$

There is a common factor to cancel:

$\frac{1-a}{a}$

The slope of AB is $\frac{1-a}{a}$ .

I'm going to use point-slope form to determine my linear equation for AB.

Point-slope form is y-y_1=m(x-x_1) where is slope and (x_1,y_1) is a point on the line that you know.

So we have that (-1,1) is the point where is $\frac{1-a}{a}$ .

Plugging this in gives us:

$y-1=\frac{1-a}{a}(x-(-1))$

$y=\frac{1-a}{a}(x+1)+1$ .

So the linear equation that goes through points A and B is:

$y=\frac{1-a}{a}(x+1)+1$ .

Now we said earlier that the line for OP will be:

$y=\text{opposite reciprocal slope of }AB \cdot x$ .

Opposite just means change the sign.

Reciprocal just means we are going to flip.

So the opposite reciprocal of $\frac{1-a}{a}$ is:

$-\frac{a}{1-a}$ .

So the equation when graphed that goes through pts O & P is:

$y=-\frac{a}{1-a}x$ .

Now to actually find this point P, I need to find the intersection of the lines I have found. This lines I found for AB & OP respectively are:

$y=\frac{1-a}{a}(x+1)+1$

$y=-\frac{a}{1-a}x$

I'm going to use substitution to solve this system.

$\frac{1-a}{a}(x+1)+1=-\frac{a}{1-a}x$

To solve this equation for x, I need to get the terms that contain x on one side while the terms not containing x on the opposing side.

I'm going to distribute the $\frac{1-a}{a}$ to both terms in the ( ) next to it:

$\frac{1-a}{a}x+\frac{1-a}{a}+1=-\frac{a}{1-a}x$

Now I'm to subtract $\frac{1-a}{a}x$ on both sides:

$\frac{1-a}{a}+1=-\frac{a}{1-a}x-\frac{1-a}{a}x$

Factor out the x on the right hand side:

$\frac{1-a}{a}+1=(-\frac{a}{1-a}-\frac{1-a}{a})x$

Now divide both sides by what x is being multiplied by:

$\frac{\frac{1-a}{a}+1}{-\frac{a}{1-a}-\frac{1-a}{a}}=x$

We need to clear the mini-fractions by multiplying top and bottom by a(1-a) :

$\frac{(1-a)^2+a(1-a)}{-a^2-(1-a)^2}$

$x=\frac{(1-a)^2+a(1-a)}{-a^2-(1-2a+a^2)}$

Distributing the - in front of the ( ) on bottom:

$x=\frac{(1-a)^2+a(1-a)}{-a^2-1+2a-a^2}$

$x=\frac{(1-a)^2+a(1-a)}{-2a^2-1+2a}$

I'm going to factor the (1-a) out on top since both of those terms have that as a common factor:

$x=\frac{(1-a)(1-a+a)}{-2a^2-1+2a}$

This simplify to:

$x=\frac{1-a}{-2a^2+2a-1}$

Now let's find the corresponding y-coord using either of one our equations.

I prefer the line for OP:

$y=-\frac{a}{1-a}x$ with $x=\frac{1-a}{-2a^2+2a-1}$

$y=-\frac{a}{1-a}(\frac{1-a}{-2a^2+2a-1})$

1-a 's cancel:

$y=-\frac{a}{-2a^2+2a-1}$

$y=\frac{a}{2a^2-2a+1}$ .

So point P is $(\frac{1-a}{-2a^2+2a-1},\frac{a}{2a^2-2a+1})$ .

So now we can actually use the distance formula to compute the thing we called the height of the triangle which was the distance between O & P:

$\sqrt{(\frac{1-a}{-2a^2+2a-1}-0)^2+(\frac{a}{2a^2-2a+1}-0)^2}$

$\sqrt{(\frac{1-a}{-2a^2+2a-1})^2+(\frac{a}{2a^2-2a+1})^2}$

$\sqrt{(\frac{-1+a}{2a^2-2a+1})^2+(\frac{a}{2a^2-2a+1})^2}$

$\sqrt{\frac{(-1+a)^2+a^2}{(2a^2-2a+1)^2}}$

Let's simplify the top using (a+b)^2=a^2+2ab+b^2 :

$\sqrt{\frac{1-2a+a^2+a^2}{(2a^2-2a+1)^2}}$

$\sqrt{\frac{1-2a+2a^2}{(2a^2-2a+1)^2}}$

$\sqrt{\frac{1}{2a^2-2a+1}}$ .

$\frac{1}{\sqrt{2a^2-2a+1}}$

Answer from: Quest

okay wait before i answer, is this for your points on a graph?

Answer from: Quest

answer: plz put a pic in so i can answer it

step-by-step explanation:

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