To answer this question, we need to find the midpoint of .

To find how far the midpoint of a line is from each end, we use the following equation:

 and  are taken from the  value of the second point and  and  are taken from the  value of the first point. Therefore, for this data:

We can then solve:

Therefore, our midpoint is  units between each endpoint's  value and  unit between each endpoint's  value. To find out the location of the midpoint, we subtract the midpoint distance from the  point. (In this case it's the point .) Therefore:

So the midpoint is located at 

The question asked us what the -coordinate of this point was. Therefore, our answer is .

The first step is to find the y-coordinates for the two points we are using. To do this we plug our x-values into the equation. Where , we get , giving us the point . Where , we get , giving us the point .

We can now use the distance formula: .

Plugging in our points gives us 

To answer this question, we must put the equation into slope-intercept form, meaning we must solve for . Slope-intercept form follows the format  where  is the slope and  is the intercept.

Therefore, we must solve the equation so that  is by itself. First we add  to both sides so that we can start to get  by itself:

Then, we must subtract  from both sides:

We then must divide each side by 

Therefore, the slope-intercept form of the original equation is .

In this kind of problem, it's important to keep track of information given about your line of interest. In this case, the coordinates given set up the stage for us to be able to get to our line of focus - the line perpendicular to the tangent line. In order to determine the perpendicular line's slope, the tangent line's slope must be calculated. Keeping in mind that:

where y2,x2 and y1,x1 are assigned arbitrarily as long as the order of assignment is maintained. 

  which is the slope of the tangent line.

To calculate the perpendicular line, we have to remember that the product of the tangent slope and the perpendicular slope will equal -1.

, the perpendicular slope can then be calculated as 

To find the slope of the tangent line, we must take the derivative.

By using the Power Rule we will be able to find the derivative:

Therefore derivative of  is .

Now we plug in , giving us .

The only two bits of information that are given for the tangent line is that it runs through the points  and . With these two points, the line's slope can be easily calculated through the equation: 

where  is slope,  is the -coordinate of the points, and  is the -coordinates of the points. 

Slope can be calculated through substituting in for the given values:

The line must be perpendicular to the radius at the point (–3,4). The slope of the radius is given by  

The radius has endpoints (–3,4) and the center of the circle (0,0), so its slope is –4/3.

The slope of the tangent line must be perpendicular to the slope of the radius, so the slope of the line is ¾.

The equation of the line is y – 4 = (3/4)(x – (–3))

Rearranging gives us: 3x – 4y = -25

The graph of the equation  is a circle with center .

A tangent to this circle at a given point is perpendicular to the radius to that point. The radius with endpoints  and  will have slope

,

so the tangent line has the opposite of the reciprocal of this, or , as its slope. 

The tangent line therefore has equation

Rewrite the equation of the circle in standard form to find its center:

Complete the square:

The center is . 

A tangent to this circle at a given point is perpendicular to the radius to that point. The radius with endpoints  and  will have slope

,

so the tangent line has the opposite of the reciprocal of this, or , as its slope. 

The tangent line therefore has equation

To find an equation tangent to

we need to find the first derviative of this equation with respect to  to get the slope  of the tangent line.

So,

due to power rule .

First we need to find our slope by plugging our  into the derivative equation and solving.

Thus, the slope is

.

To find the equation of a tangent line of a given point  we plug the point into

.

Therefore our equation becomes,

Once we rearrange, the equation is