# Matrix transformation shearing the triangle to be right triangle

If we have a triangle at $(1,1), (5,3), (7,1)$, how to find the sheared matrix to transform the triangle to be right triangle at $(1,1)$.

Is it that we need to find $i$ in
$\begin{pmatrix} 1 && i \\ 0 && 1 \\ \end{pmatrix}$

But how to ensure the right triangle.

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Definition

Coordinate Transformation

Given a 2D Point $(x,y)$, to transform the point to another 2D coordinate space is defined through the following set of equations $$x'=a\cdot x + b\cdot y$$ $$y'=c\cdot c + d\cdot y$$

where $a,b,c,d$ are real value constant with the constraint $a\cdot d - b\cdot c \ne 0$

And can be written in the matrix form as

$$\begin{pmatrix} x' \\ y' \\ \end{pmatrix} = \begin{pmatrix} a && b \\ c && d \\ \end{pmatrix} \begin{pmatrix} x \\ y \\ \end{pmatrix}$$ or in short $$T( \vec x ) = \mathbf{A} \vec x$$ where $\vec x$ is a column vector. For multiple vectors, the same can be written as $$T( \vec X ) = \mathbf{A} \vec X \tag1$$ where $$T(\vec X) = \left [\vec x_1\, \vec x_2\, ....\, \vec x_n \right ]$$ In your case $$A=\begin{pmatrix} 1 && i \\ 0 && 1 \\ \end{pmatrix}$$ $$X=\begin{pmatrix} 1 && 5 && 7 \\ 1 && 3 && 1\\ \end{pmatrix}$$

Representing as $(1)$ we get $$T(\vec X)=\begin{pmatrix} 1 && i \\ 0 && 1 \\ \end{pmatrix} \begin{pmatrix} 1 && 5 && 7 \\ 1 && 3 && 1\\ \end{pmatrix}$$ $$\Rightarrow T(\vec X)=\begin{pmatrix}i + 1 & 3 i + 5 & i + 3\\1 & 3 & 1\end{pmatrix}\tag2$$ $$\Rightarrow T(\vec X)=\left [ \vec x_1\,\vec x_2\,\vec x_3\right ]$$ Once we determine the transposed vector, we need to determine the dot product of the respective vectors to determine, which of them equates to 0 satisfy the condition where in one of the points is $(1,1)$ $$\vec x_1 \cdot \vec x_2 = \left(i + 1\right) \left(3 i + 5\right) + 3$$ Solving which gives $$\begin{bmatrix}- \frac{4}{3} - \frac{2}{3} \sqrt{2} \mathbf{\imath}, & - \frac{4}{3} + \frac{2}{3} \sqrt{2} \mathbf{\imath}\end{bmatrix}$$ $$\vec x_2 \cdot \vec x_3 = \left(i + 3\right) \left(3 i + 5\right) + 3$$ Solving which gives \begin{bmatrix}- \frac{7}{3} - \frac{1}{3} \sqrt{5} \mathbf{\imath}, & - \frac{7}{3} + \frac{1}{3} \sqrt{5} \mathbf{\imath}\end{bmatrix} $$\vec x_3 \cdot \vec x_1 = \left(i + 1\right) \left(i + 3\right) + 1$$ Solving which gives \begin{bmatrix}-2\end{bmatrix} Replacing each of these values in $(2)$, its evident that only the last solution satisfies the conditions \begin{pmatrix}-1 & -1 & 1\\1 & 3 & 1\end{pmatrix}

So $i=-2$ is the solution to the problem

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