How do you find the implicit form $(Ax + By + C = 0)$ in linear algebra, given this example?

Question

I have a point $(x_0, y_0)$ at $(1, 2)$ and a directional vector $\vec{v}=\left[\frac{3}{5} \frac{4}{5}\right]$. The corresponding implicit form is $$4x - 3y + 2 = 0$$ How did they derive this using this information?

Answer 1

In general: Let the point be $(x_0, y_0)$ and the vector $v = [a\; b]$. The implicit form is:

$$-b(x - x_0) + a(y - y_0) = 0. \quad (*)$$

The derivation is simple: the slope of $Ax + By + c = 0$ is given by $-A/B$ (take care of $B = 0$ separately).
Moreover, given the slope of a line and a point on it, the line is uniquely determined. Now, you may check that $(*)$ is indeed a line which has inclination in the direction of $[a\; b]$ which passes through $(x_0, y_0)$ and hence, it is the desired line.

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In your case, we have $$-4/5(x - 1) + 3/5(y - 2) = 0$$ which simplifies to $$-4x + 3y - 2 = 0$$ or the thing which you have written.

Answer 2

The vector equation of the line will be $$\vec r= \hat i + 2\hat j + t\left( \frac 35 \hat i + \frac 45 \hat j\right)$$

Write $r$ as $x\hat i + y\hat j$ and compare to get

$$x=1+\frac 35 t \\ y=2+\frac 45 t \implies \frac t5 = \frac{y-2}{4}$$

Finally, substitute this value in the first equation

$$x=1+\frac 34 (y-2) \\ \implies 4x-3y+2=0$$