Given two points $A = [-2, 3]$ and $B = [3; -1]$, find the equation of the line $p =AB\ $in its general form.
Joseph's solution:
(1) The general form of the equation of a straight line is $ax + by + c = 0$.
(2) Since the points $A$ and $B$ lie on the line $p$, the direction vector of the line $p$ is $\ \overrightarrow{u}=\overrightarrow{AB}= B\ –\ A = (5; -4)$.
(3) The direction vector $\ \overrightarrow{u}=(a;b)$, therefore, the equation of $p$ is $5x\ –\ 4y + c = 0$.
(4) Point $A$ lies on the line $p$, so $5\cdot(-2)\ –\ 4 \cdot 3 + c = 0 \Rightarrow c = 22$.
(5) The equation of the line $p$ in general form is $5x\ –\ 4y + 22 = 0$.
Is Joseph's solution incorrect? If so, where did Joseph make a mistake in his procedure?
Joseph's solution is correct.
The mistake is in step (2). The direction vector of the line $p$ is $\ \overrightarrow{u}=\overrightarrow{AB}=B\ –\ A=(1; -4)$.
The mistake is in step (3). The direction vector $\ \overrightarrow{u}\neq (a;b)$.
The mistake is in step (4). $A\in p$ and so $5\cdot(-2)\ –\ 4\cdot3+c=0\Rightarrow c = -22$.
(1) The general form of the equation of a straight line is $ax + by + c = 0$.
(2) Since the points $A$ and $B$ lie on the line $p$, the direction vector of the line $p$ is $\ \overrightarrow{u}=\overrightarrow{AB}=B\ –\ A=(5; -4)$.
(3) The coefficients $a$ and $b$ in the general equation of the line are the components of the normal vector (not the direction vector). It is known that $\ \overrightarrow{n}\ \bot\ \overrightarrow{u}$, so $\overrightarrow{n}=(4;5)$. Then the equation of $p$ is $4x + 5y + c = 0$.
(4) Point $A$ lies on the line $p$, so $4\cdot(-2) + 5 \cdot 3 + c = 0 \Rightarrow c = -7$.
(5) The equation of the line $p$ in general form is $4x + 5y\ –\ 7 = 0$.