# how do find the line lies on another line

I have Line $A$ with points $(1,1)$ and $(8,8)$ and another Line $B$ with points $(2,2)$ and $(4,4)$. I would like to prove Line $B$ lies on Line $A$. How can I do this?

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The points are all on the line $y=x$.

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In a general version of your situation, with one line determined by points $A$ and $B$ and a second line determined by points $C$ and $D$, the idea from anonymous's answer is a good place to start. Rather than using the quotient formulation in that answer, which might have division by zero, let's rewrite that with products: $$(x-x_1)(y_2-y_1)=(y-y_1)(x_2-x_1)$$ Now, use that to write an equation for the first line, through $A(x_1,y_1)$ and $B(x_2,y_2)$. Given that equation, any points $(x,y)$ that satisfy the equation are on the line. Test points $C$ and $D$ in that equation. If both $C$ and $D$ satisfy the equation for the line through $A$ and $B$, then all four points are on the same line, so the line through $C$ and $D$ is the same as the line through $A$ and $B$.

In this way, we avoid having to write a second equation and determine whether or not that equation is equivalent to the first equation; instead, we look to see if the points that would determine the second equation satisfy the first equation.

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Hint: Take the perpendicular distance and prove that the distance is 0.

Use the 2 point slope form.Equation of the line passing through $(x_{1},y_{1})$ and $(x_{2},y_{2})$ is given bye $$\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}$$ and observe that they are same.

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In general you can use a determinant expression

$$\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}=0$$

to verify that three points $(x_i,y_i)$, $i=1\dots 3$ are collinear. As Hans points out though, that the coordinates of all four points that you have have their abscissas and ordinates identical makes them obviously collinear.

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