I have these vertices of a triangle
A 5,20
B 5,30
C 15, 25
What are the equations of each of the three line that is: AB, BC, and AC?
What is the length of each of the three lines?
Thanks very much.
[Math] triangle, calculate the length using the vertices.
geometry
Related Solutions
Let us derive a formula for rotating a point $(x_1,y_1)$ clockwise about another point $(x_0,y_0)$.
First we translate the point so that the point $(x_0,y_0)$ becomes the origin. We do this by subtracting $x_0$ from the $x$ coordinate of all points, and subtracting $y_0$ from the $y$ coordinate of all points.
So now our two point become $(x_1-x_0,y_1-y_0)$ and the origin.
Now let's shift our attention to something else what happens when your rotate the unit vector $\vec i=\langle 1,0 \rangle$ clockwise $\theta$ degrees about the origin.
We get the vector $\langle \cos (\theta), -\sin (\theta) \rangle$.
What happens when we rotate the vector $\vec j=\langle 0,1 \rangle$ in the same way. We get the vector $\langle \sin (\theta), \cos (\theta) \rangle$.
Now why are we interested in these two vectors? Well because, it turns out that rotation in two dimensions clockwise about the origin is a linear operator. Meaning that if $x_1-x_0$ and $y_1-y_0$ are constants we have,
$$\text{Rot}_{\theta}((x_1-x_0) \vec V+(y_1-y_0) \vec W)=(x_1-x_0)\text{Rot}_{\theta} \vec V+(y_1-y_0) \text{Rot}_{\theta} \vec W$$
Now we write the point we want to rotate, that being $(x_1-x_0,y_1-y_0)$ as a vector $\langle x_1-x_0, y_1-y_0 \rangle=(x_1-x_0) \vec i+(y_1-y_0) \vec j $
And we rotate noting the $\text{Rot}$ function is a linear operator.
$$\text{Rot}_{\theta}((x_1-x_0) \vec i+(y_1-y_0) \vec j )$$
$$=(x_1-x_0)\text{Rot}_{\theta} \vec i+(y_1-y_0) \text{Rot}_{\theta} \vec j$$
$$=(x_1-x_0) \langle \cos (\theta), -\sin (\theta) \rangle+(y_1-y_0)\langle \sin (\theta), \cos (\theta) \rangle$$
$$=\langle (x_1-x_0) \cos (\theta)+(y_1-y_0) \sin (\theta), -(x_1-x_0)\sin (\theta)+(y_1-y_0) \cos (\theta) \rangle$$
Finally we undo the translation we did in the begging by adding $x_0$ and $y_0$ to the $x$ and $y$ coordinate respectively to get.
$$\left((x_1-x_0) \cos (\theta)+(y_1-y_0) \sin (\theta)+x_0, -(x_1-x_0)\sin (\theta)+(y_1-y_0) \cos (\theta)+y_0 \right)$$
Rotating $A$ gives,
$$\left((5-\frac{25}{3})\frac{\sqrt{2}}{2}+(20-25) \frac{\sqrt{2}}{2}+\frac{25}{3}, -(5-\frac{25}{3})\frac{\sqrt{2}}{2}+(20-25) \frac{\sqrt{2}}{2}+25 \right)$$
This is approximately,
$$(2.4408,23.8215)$$
Using the construction that @Michael Rozenberg suggested
I will leave the following exercise for you (which isn't that hard)
Prove that the quadrilateral $ABCD$ is cyclic.
Thus $\angle BDC=180°-\angle CAB=120°$. In virtue of the law of Cosines $$\begin{array}a [CB]^2&=[CD]^2+[DB]^2-2·[CD]·[DB]·\cos(\angle BDC)\\ &=1+4-2·1·2·(-0.5)\\ &=5+2=7 \end{array}$$
Best Answer
Hint -
For equations use following formula-
$(y - y_1) = \frac{(y_2 - y_1)}{(x_2 - x_1)} . (x - x_1)$
For length -
$\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$