How to find the coordinate of $B'$ and $C'$ in an efficient way?

Question

where $B'C'$ is parallel to $BC$ and the distance between them is $d$. I have the coordinate of $A, B,$ and $C$ and the value of $d$;

Answer 1

Draw the perpendicular from $A$ to $BC$ and $B'C'$. Let's call the intersection points $D$ and $D'$. The length of $AD$ is $h$. Then using similar triangles you can get $$B'=B+(A-B)\frac{|DD'|}{|AD|}=B+(A-B)\frac dh$$ and similar for $C'$. In order to get the height $h$, I would use the area of the triangle written as $\frac 12 |BC|h$ and using Heron's formula

Answer 2

Say $\beta$ the angle $ABC=ACB$, I have: $\beta=\tan^{-1}(\frac{2y_A}{x_C})$. $AB'=C'C=\frac{d}{\sin(\beta)}$. The points are: $$B'(d\cdot \frac{\cos(\beta)}{\sin(\beta)},d)$$ and $$C'(x_c-d\cdot \frac{\cos(\beta)}{\sin(\beta)},d)$$.

I have taken for simplicity $B(0,0)$.

Answer 3

You can find the equation of the line that passes through $B'$ and $C'$ and find its interception with the line that passes through $AB$ and $AC$, since $BC$ is parallel to $B'C'$, their slopes are the same, so you need at least one point through that line, if the base of the triangle is parallel to the $x-axis$, and the coordinates of $B$ are $(x_1, y_1)$ then a point on that line would be $(x_1, y_1 + d)$, if the base is parallel to the $y-axis$, then a point would be $(x_1 + d, y_1)$ and if they are not parallel to any axis, you can use Pythagorean theorem to find that point, once you have it, just find the equation of the line betwen $B'C'$, $AB$ and $AC$ and find their interception.

Answer 4

At first find scale factor of similarity of triangles $ABC, AB{'}C{'};$

$$p<1= \frac {AB^{'}}{AB}$$

for all sides $(c,a,b)$ and altitude $h$.

$$ b \sin C\, ( = c \sin B) = h;\,p b \sin C = ph $$

Subtracting

$$ h(1-p) = d=b \sin C \, (1-p) \tag1 $$

Calculate $p$ since $(d,b,C) $ are known,

$$ p= 1- \frac{d}{b \sin C};$$

Next using proportional division of each side we have

$$ x_{B^{'}} = p\, x_{A} + (1-p) x_{B}$$ $$ y_{B^{'}} = p\, y_{A} + (1-p) y_{B}$$

Similarly

$$ x_{C^{'}} = p\, x_{A} + (1-p) x_{C}$$ $$ y_{C^{'}} = p\, y_{A} + (1-p) y_{C}.$$

Answer 5

If $d\perp BC$, then for example with vectors:

$$B'=(x_B+d\cdot \vec{BA}\cdot\vec{BC}/\sqrt{(|\vec{BA}||\vec{BC}|)^2-(\vec{BA} \cdot\vec{BC})^2},y_B+d)$$

$$C'=(x_C-d\cdot \vec{CA}\cdot\vec{CB}/\sqrt{(|\vec{CA}||\vec{CB}|)^2-(\vec{CA} \cdot\vec{CB})^2},y_C+d)$$

or you can write out the line equations and find intersection points $B', C'$ or use the similarity of $\Delta ABC$ and $\Delta AB'C'$.

You can calculate it step by step as follows:

First find the angle $\theta_B$ at point $B$ between the vectors $\vec{BA}$ and $\vec{BC}$. This can be done using the dot product. First build the vectors and calculate their lengths:

1. $\vec{BA}$ between the point $B=(x_B,y_B)$ and the point $A=(x_A,x_A)$:

$$\vec{BA}=\begin{bmatrix}x_A-x_B\\y_A-y_B\end{bmatrix}, \ |\vec{BA}|=\sqrt{(x_A-x_B)^2+(y_A-y_B)^2}$$

2. $\vec{BC}$ between the point $B=(x_B,x_B)$ and the point $C=(x_C,y_C)$:

$$\vec{BC}=\begin{bmatrix}x_C-x_B\\y_C-y_B\end{bmatrix}, \ |\vec{BC}|=\sqrt{(x_C-x_B)^2+(y_C-y_B)^2}$$

then use the formula of the dot product to get $\cos(\theta_B)$: $$\cos(\theta_B)=\frac{\vec{BA} \cdot \vec{BC} }{ |\vec{BA}|\cdot |\vec{BC}|} = \frac{(x_A-x_B)\cdot (x_C-x_B)+(y_A-y_B)\cdot (y_C-y_B)}{\sqrt{(x_A-x_B)^2+(y_A-y_B)^2}\cdot\sqrt{(x_C-x_B)^2+(y_C-y_B)^2}}$$ and the angle $\theta_B$ at point $B$ between the vectors $\vec{BA}$ and $\vec{BC}$: $$\theta_B=\arccos\left(\frac{(x_A-x_B)\cdot (x_C-x_B)+(y_A-y_B)\cdot (y_C-y_B)}{\sqrt{(x_A-x_B)^2+(y_A-y_B)^2}\cdot\sqrt{(x_C-x_B)^2+(y_C-y_B)^2}}\right)$$ Similarly you can find the angle $\theta_C$ at point $C$ between the vectors $\vec{CB}$ and $\vec{CA}$: $$\theta_C=\arccos\left(\frac{(x_B-x_C)\cdot (x_A-x_C)+(y_B-y_C)\cdot (y_A-y_C)}{\sqrt{(x_B-x_C)^2+(y_B-y_C)^2}\cdot\sqrt{(x_A-x_C)^2+(y_A-y_C)^2}}\right)$$

A vector $\vec{d}$ with length equals $1$, which is perpendicular to $\vec{BC}$ can be given as follows: $$\vec{\hat{d}}=\frac{1}{\sqrt{(x_B-x_C)^2+(y_B-y_C)^2}}\cdot\begin{bmatrix}-(y_C-y_B)\\x_C-x_B\end{bmatrix}$$ A vector $\vec{d}_{\perp}$ with length equals $1$, which is perpendicular to $\vec{d}$ or in the same direction of $\vec{BC}$, can be given as follows: $$\vec{\hat{d}_{\perp}}=\frac{1}{\sqrt{(x_B-x_C)^2+(y_B-y_C)^2}}\cdot\begin{bmatrix}x_C-x_B\\y_C-y_B\end{bmatrix}$$ Then by vector addition and with the distance $d$ between the line segment $BC$ and the line segment $B'C'$ we get $$\vec{B'} =\vec{B}+d\cdot \vec{\hat{d}}+\left(\frac{d}{\tan(\theta_B)}\right)\cdot\vec{\hat{d}_{\perp}} $$

and

$$\vec{C'} =\vec{C}+d\cdot \vec{\hat{d}}-\left(\frac{d}{\tan(\theta_C)}\right)\cdot\vec{\hat{d}_{\perp}} $$

Here is a worked example:

Assume $$A=\begin{bmatrix}x_A\\y_A\end{bmatrix}=\begin{bmatrix}0\\5\end{bmatrix},~B=\begin{bmatrix}x_B\\y_B\end{bmatrix}=\begin{bmatrix}-5\\0\end{bmatrix},~C=\begin{bmatrix}x_C\\y_C\end{bmatrix}=\begin{bmatrix}5\\0\end{bmatrix},~d=2$$ then: $$\theta_B=\arccos\left(\frac{(0-(-5))\cdot (5-(-5))+(0-0)\cdot (0-0)}{\sqrt{(0-(-5))^2+(5-0)^2}\cdot\sqrt{(5-(-5))^2+(0-0)^2}}\right)=\arccos\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4}$$ and $$\theta_C=\arccos\left(\frac{((-5)-5)\cdot (0-5)+(0-0)\cdot (5-0)}{\sqrt{((-5)-5)^2+(0-0)^2}\cdot\sqrt{(0-5)^2+(5-0)^2}}\right)=\arccos\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4}$$ and $$\vec{\hat{d}}=\frac{1}{\sqrt{((-5)-5)^2+(0-0)^2}}\cdot\begin{bmatrix}-(0-0)\\5-(-5)\end{bmatrix}=\begin{bmatrix}0\\1\end{bmatrix}$$ and $$\vec{\hat{d}_{\perp}}=\frac{1}{\sqrt{((-5)-5)^2+(0-0)^2}}\cdot\begin{bmatrix}5-(-5)\\0-0\end{bmatrix}=\begin{bmatrix}1\\0\end{bmatrix}$$ and $$\tan(\theta_B)=\tan(\theta_C)=\tan\left(\frac{\pi}{4}\right)=1$$ So $$B'=\begin{bmatrix}-5\\0\end{bmatrix}+2\cdot \begin{bmatrix}0\\1\end{bmatrix}+2\cdot\begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}-3\\2\end{bmatrix} $$ and $$C'=\begin{bmatrix}5\\0\end{bmatrix}+2\cdot \begin{bmatrix}0\\1\end{bmatrix}-2\cdot\begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}3\\2\end{bmatrix}$$

We also could develop the line equations and solve for the intersection points, we get the following solutions: $$B' = \frac{y_A-y_C-d}{y_A-y_C}B+\frac{d}{y_A-y_C}A = \frac{y_A-y_C-d}{y_A-y_C}\begin{bmatrix}x_B\\y_B\end{bmatrix}+\frac{d}{y_A-y_C}\begin{bmatrix}x_A\\y_A\end{bmatrix}$$ and $$ C' = \frac{y_C-y_A+d}{2y_C-y_A-y_B}C+\frac{y_C-y_B-d}{2y_C-y_A-y_B}A = \frac{y_C-y_A+d}{2y_C-y_A-y_B}\begin{bmatrix}x_C\\y_C\end{bmatrix}+\frac{y_C-y_B-d}{2y_C-y_A-y_B}\begin{bmatrix}x_A\\y_A\end{bmatrix} $$ Evaluating for the above example: $$B' = \frac{5-0-2}{5-0}\begin{bmatrix}-5\\0\end{bmatrix}+\frac{2}{5-0}\begin{bmatrix}0\\5\end{bmatrix}=\begin{bmatrix}(3/5)\cdot (-5)\\(3/5)\cdot 0\end{bmatrix}+\begin{bmatrix}(2/5)\cdot 0\\(2/5)\cdot 5\end{bmatrix}=\begin{bmatrix}-3\\2\end{bmatrix} $$ and $$ C' = \frac{0-5+2}{ 0-5-0}\begin{bmatrix}5\\0\end{bmatrix}+\frac{0-0-2}{ 0-5-0}\begin{bmatrix}0\\5\end{bmatrix}=\begin{bmatrix}((-3)/(-5))\cdot 5\\((-3)/(-5))\cdot 0\end{bmatrix}+\begin{bmatrix}((-2)/(-5))\cdot 0\\((-2)/(-5))\cdot 5\end{bmatrix}=\begin{bmatrix}3\\2\end{bmatrix} $$

Answer 6

$\angle ABC = \beta = tan^{-1}(m_{AB}) – tan^{-1}(m_{BC})$ = a known quantity.

Let B’ divide BA in the ratio 1 : r.

Let the circle using AB as diameter cut BC at P. $AP = D = AB \sin \beta$, a therefore known quantity.

$\dfrac {D}{d} = \dfrac {1 + r}{1}$. That is, r : 1 = (D – d) : d

Then, $x_{B’} = \dfrac {(D – d) \times x_B + d \times x_A}{D}$ by the section formula.