algebra-precalculussystems of equations
My maths TB says:
For the simultaneous equations, $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ if $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$ then the pair of equations are said to be consistent and dependent, and when plotted produce coincident lines.
I understand that these are consistent in the sense that they have some solution(though infinite).
But why are these called dependent?
Thanks in advance!
Clearly, we can scale the coefficients of a given linear equation by any (non-zero) constant and the result is unchanged. Therefore, by dividing-through by $\sqrt{a_i^2+b_i^2}$, we may assume our equations are in "normal form":
$$\begin{align} x \cos\theta + y \sin\theta - p &= 0 \\ x \cos\phi + y \sin\phi - q &= 0 \\ x \cos\psi + y \sin\psi - r &= 0 \end{align}$$
with $\theta$, $\phi$, $\psi$ and $p$, $q$, $r$ (and $A$, $B$, $C$ and $a$, $b$, $c$) as in the figure:
Then
$$C_1 = \left|\begin{array}{cc} \cos\phi & \sin\phi \\ \cos\psi & \sin\psi \end{array} \right| = \sin\psi\cos\phi - \cos\psi\sin\phi = \sin(\psi-\phi) = \sin \angle ROQ = \sin A$$ Likewise, $$C_2 = \sin B \qquad C_3 = \sin C$$
Moreover, $$D := \left|\begin{array}{ccc} \cos\theta & \sin\theta & - p \\ \cos\phi & \sin\phi & - q \\ \cos\psi & \sin\psi & - r \end{array}\right| = - \left( p C_1 + q C_2 + r C_3 \right) = - \left(\;p \sin A + q \sin B + r \sin C\;\right)$$
Writing $d$ for the circumdiameter of the triangle, the Law of Sines tells us that $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = d$$
Therefore,
$$\begin{align} D &= - \left( \frac{ap}{d} + \frac{bq}{d} + \frac{cr}{d} \right) \\[4pt] &= -\frac{1}{d}\left(\;ap + b q + c r\;\right) \\[4pt] &= -\frac{1}{d}\left(\;2|\triangle COB| + 2|\triangle AOC| + 2|\triangle BOA| \;\right) \\[4pt] &= -\frac{2\;|\triangle ABC|}{d} \end{align}$$
Also, $$C_1 C_2 C_3 = \sin A \sin B \sin C = \frac{a}{d}\frac{b}{d}\sin C= \frac{2\;|\triangle ABC|}{d^2}$$
Finally: $$\frac{D^2}{2C_1C_2C_3} = \frac{4\;|\triangle ABC|^2/d^2}{4\;|\triangle ABC|/d^2} = |\triangle ABC|$$
If two lines do not intersect--i.e., they have no points in common--then the system of equations $$\begin{align*} a_1 x + b_1 y + c_1 &= 0 \\ a_2 x + b_2 y + c_2 &= 0 \end{align*}$$ will have no solution for $(x,y)$. Thus, if we solve the system, we find $$x = \frac{b_1 c_2 - b_2 c_1}{a_1 b_2 - a_2 b_1}, \quad y = \frac{a_2 c_1 - a_1 c_2}{a_1 b_2 - a_2 b_1}.$$ This solution does not exist or is indeterminate if $a_1 b_2 - a_2 b_1 = 0$.
However, some care is required: two lines coincide if $$(a_1, b_1, c_1) = k(a_2, b_2, c_2)$$ for some nonzero scalar constant $k$, and in this case, both the numerators and denominator of the aforementioned solution are zero, meaning that there are infinitely many points that the two equations share in common. So while it is a sufficient condition for $a_1 b_2 - a_2 b_1 \ne 0$ to imply that the lines intersect, it is not a strictly necessary condition, and for that reason, the question should have been better phrased by saying "two lines...intersect if..." rather than "only if"; alternatively, it might be phrased "two distinct lines...intersect only if...."
Best Answer
An equation $R(x_1,x_2, ... , x_n)=0$ is called dependent on an equation $S(x_1, x_2, ..., x_n)=0$ if the equation $R(x_1,x_2, ... , x_n)=0$ can be derived algebraically (by using a finite number of the operations of addition, subtraction, multiplication, division, and exponentiation with constant rational exponents) from the equation $S(x_1, x_2, ..., x_n)=0$. If two equations are dependent on each other, we call each of them dependent.
According to the Merriam-Webster dictionary, dependent means
So, calling such equations dependent seems reasonable because knowing each one can determine the other.
For example, let $R(x,y)=a_1x+b_1y+c_1=0$ and $S(x,y)=a_2x+b_2y+c_2=0$. Suppose that $k=\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$ is a nonzero constant. Then we can derive $R(x,y)=0$ from $S(x,y)=0$ by multiplying both sides of $S(x,y)=0$ by $k$ as follows.$$k(a_2x+b_2y+c_2)=k(0) \quad \Rightarrow \quad a_1x+b_1y+c_1=0.$$Similarly, we can derive $S(x,y)=0$ from $R(x,y)=0$ by dividing both sides of $S(x,y)=0$ by $k$ as follows.$$\frac{a_1x+b_1y+c_1}{k}=\frac{0}{k} \quad \Rightarrow \quad a_2x+b_2y+c_2=0.$$Thus, the equations $R(x,y)=0$ and $S(x,y)=0$ are dependent.
Addendum
Please note that the dependence of two equations need not to be linear; however, if it is linear, like your example, then the solution set of the equations are the same.