Finding Coordinates of a Point That Creates a Right Angle The question is as follows: 

Let $A$ = $(0, 0)$ and $C$ = $(4, 3)$. Point $D$ is located so that angle $ACD$ is a right angle and the tangent of angle $DAC$ is 3/4. Find coordinates for $D$. There are two answers. 

I was just able to find the slope of AC, which is 3/4. So that means that the slope perpendicular to that would be $-\frac{4}{3}$. But I don't know how to go any further than that and I also am unsure about how to use the tangent of angle $DAC$ for this problem. Any help will be greatly appreciated.
 A: The slope of $AC$ has been calculated wrong : it is in fact $\frac{3-0}{4-0} = \frac 34$. Hence, the slope of the line perpendicular to $AC$ must be $\frac{-4}{3}$. So any point $D$ is located on the straight line passing through $C$ and having slope $\frac{-4}{3}$ i.e. $(y-3) = -\frac 43(x-4)$. This simplifies to $3y+4x = 25$. 
Hence, $D$ satisfies these coordinates. Furthermore, the length of $AC$ is $5$, as you have said above. 
Try to imagine the triangle ACD in your mind. This triangle is right angled at $C$, and the length of $AC$ is $5$.
EDIT : What is the definition of tangent of $DAC$? It is defined as the ratio between the opposite and adjacent sides, right? That is, $\tan DAC = \frac{OPP}{ADJ}$, where $OPP$ is the side opposite the angle $DAC$, and $ADJ$ is the side adjacent to $DAC$ which is not the hypotenuse (for that, we use the phrase $HYP$, so this side is not the hypotenuse).
Picturing the triangle in your mind, you must be able to see that $OPP = DC$, and $ADJ = AC$. Hence, $\tan DAC = \frac{DC}{AC} = \frac 34$. Therefore, $DC = \frac{3AC}{4} = \frac{3 \times 5}{4} = \frac{15}{4} = 3.75$.
Now, you have a right angled triangle, hence by Pythagoras' theorem, $AD^2 = AC^2+DC^2$ (from your diagram, you must have seen that $AD$ is opposite angle $C$ which is right angled, so it is the hypotenuse). From this, $AD^2 = 25 + 14.0625 = 39.0625$, hence $AD = 6.25$. 
So, if $D= (x,y)$, then $x^2+y^2 = 39.0625$ by the distance from $0$ being $0.625$, and $3x+4y = 25$. Solving, you get two points : $(x,y) = (0,6.25)$, and $(x,y) = (6,1.75)$. These are the two candidate points for $D$.
POST-EDIT: If now you have understood, then I am willing to clarify two things. One, you might be wondering if the calculation to find $AD = 6.25$ is difficult. After all, we were (doing the equivalent of) squaring $3$ digit numbers there, and some six-digit arithmetic. But in fact, that part is really easy, and I can tell you why.
Secondly, how did I solve the equations $x^2+y^2 = 39.0625$ and $3x+4y=25$? That too was simple,and I can explain that as well if you have not understood yet.
POST-POST-EDIT : To solve this equation, we eliminate $y$ via $4y = 25-3x$. Multiplying the first equation by sixteen gives $16x^2+16y^2 = 625$, so $16x^2+(25-3x)^2 = 625$, a quadratic equation in $x$. From here it should be easy.
A: Hint:)
Since slope of $AC$ is $\dfrac34$ and $\tan DAC=\dfrac34$, use $\tan(x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}$ for finding the slope of $AD$. Find the intersection of lines $AD$ and $DC$.
A: You should sketch it:
$\hspace{3cm}$
Using the distance formula, you can find:
$$AC=\sqrt{(4-0)^2+(3-0)^2}=5 \ \ \text{(which is basically Pythagoras formula)}.$$
Given $\tan \angle D_1AC=\tan \angle D_2AC=\frac34$, you can find:
$$\tan \angle D_1AC=\frac{CD_1}{AC}=\frac34 \Rightarrow CD_1=\frac{3AC}{4}=\frac{15}4=CD_2.$$
Note that the triangles $ABC$ and $BCE$ are similar, hence:
$$\frac{CE}{AC}=\frac{BC}{AB} \Rightarrow \frac{CE}{5}=\frac{3}{4} \Rightarrow CE=\frac{15}{4}=CD_2 \Rightarrow E=D_2;\\
\frac{BE}{BC}=\frac{BC}{AB} \Rightarrow BE=\frac{BC^2}{AB}=\frac{9}{4} \Rightarrow D_2\left(\frac{25}{4},0\right)$$
Now you can use the similarity of the triangles $BCE$ and $FD_1E$ to find the coordinates of $D_1$. It is an exercise for you. Answer is:

$D_1\left(\frac74,6\right)$

