Question

Here is the question again, thanks for your help.

Answer 1

This is the one I referred to as Problem 3.

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:

`(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
`((1)/(a),(1)/(a^2))`

So entering these points into the formula gives us:

`((1)/(a ^2)-1)/((1)/(a)-(-1))`

Simplifying:

`((1)/(a^2)-1)/((1)/(a)+1)`

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

`(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
`a` so I can just factor that out of the two terms.

Let's do that:

`((1-a)(1+a))/(a(1+a))`

There is a common factor to cancel:

`(1-a)/(a)`

The slope of AB is
`(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
`m` 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
`m` is
`(1-a)/(a)`.

Plugging this in gives us:

`y-1=(1-a)/(a)(x-(-1))`

`y=(1-a)/(a)(x+1)+1`.

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

`y=(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
`(1-a)/(a)` is:

`-(a)/(1-a)`.

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

`y=-(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=(1-a)/(a)(x+1)+1`

`y=-(a)/(1-a)x`

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

`(1-a)/(a)(x+1)+1=-(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
`(1-a)/(a)` to both terms in the ( ) next to it:

`(1-a)/(a)x+(1-a)/(a)+1=-(a)/(1-a)x`

Now I'm to subtract
`(1-a)/(a)x` on both sides:

`(1-a)/(a)+1=-(a)/(1-a)x-(1-a)/(a)x`

Factor out the x on the right hand side:

`(1-a)/(a)+1=(-(a)/(1-a)-(1-a)/(a))x`

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

`((1-a)/(a)+1)/(-(a)/(1-a)-(1-a)/(a))=x`

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

`((1-a)^2+a(1-a))/(-a^2-(1-a)^2)`

`x=((1-a)^2+a(1-a))/(-a^2-(1-2a+a^2))`

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

`x=((1-a)^2+a(1-a))/(-a^2-1+2a-a^2)`

`x=((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=((1-a)(1-a+a))/(-2a^2-1+2a)`

This simplify to:

`x=(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=-(a)/(1-a)x` with
`x=(1-a)/(-2a^2+2a-1)`

`y=-(a)/(1-a)((1-a)/(-2a^2+2a-1))`

`1-a`'s cancel:

`y=-(a)/(-2a^2+2a-1)`

`y=(a)/(2a^2-2a+1)`.

So point P is
`((1-a)/(-2a^2+2a-1),(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{((1-a)/(-2a^2+2a-1)-0)^2+((a)/(2a^2-2a+1)-0)^2}`

`\sqrt{((1-a)/(-2a^2+2a-1))^2+((a)/(2a^2-2a+1))^2}`

`\sqrt{((-1+a)/(2a^2-2a+1))^2+((a)/(2a^2-2a+1))^2}`

`\sqrt{((-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{(1-2a+a^2+a^2)/((2a^2-2a+1)^2)}`

`\sqrt{(1-2a+2a^2)/((2a^2-2a+1)^2)}`

`\sqrt{(1)/(2a^2-2a+1)}`.

`(1)/(√(2a^2-2a+1))`