Lorentz Transformation Geometric Interpretation

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So, I was recently trying to understand special relativity and from my understanding the Lorentz transformation can be framed mathematically as follows (we are assuming a reference frame moving at speed $v$ in direction of $x$)

Let $O=(0,0)$ of a 2-D co-ordinate system

Let a line $t'$ be defined as $y=\frac{c}{v}*x$ and a line $x'$ be defined as $y=\frac{v}{c}*x$

Now consider a point $P=(x_0,y_0)$

Through $P$ we draw a line $l$ parallel to $t'$ and let $Q=l\cap x'$

Then the new co-ordinate $x_0'$ is the distance OP along $x'$

So, the formula for the line $l$ is $y-y_0=\frac{c}{v}*(x-x_0)$

So, $y=\frac{c}{v}*(x-x_0)+y_0$

We equate for $Q\in x'$

Now, we know that $Q_y=\frac{c}{v}*(Q_x-x_0)+y_0=\frac{v}{c}*Q_x$ , an equation which can be solved for $Q_x$ and $Q_y$

We know also that $OQ^2=Q_x^2+Q_y^2=Q_x^2(1+v^2/c^2)$

However evaluating this after solving yields $OQ=\frac{cx_0-vy_0}{c^2-v^2}*\sqrt{(c^2+v^2)}$

This yields $x'=\gamma(x-vt)*k$ where $k=\sqrt{(1-v^4/c^4)}$

Where on Earth did this extra factor come from?