For what value of $K$ are the points $A(1, -1)$, $B(K/2, K/3)$, and $C(4, 1)$ collinear?

Question

For what value of $K$ are the points $A(1, -1)$, $B(K/2, K/3)$, and $C(4, 1)$ collinear?

The equation of line $AC$ is $2x-3y=5$. But the point $B$ doesn't satisfy the equation.

Answer 1

"The equation of line AC is 2x-3y=5. But the point B doesnt satisfy the equation."

WHy not?

If that is the equation of the line and if we are told that $(\frac K2, \frac K3)$ is a point of the line, then it must be true that $2*\frac K2 - 3*\frac K3 = 5$. For what value of $K$ is that true?

1) Is that the equation of the line?

The slope of the equation of the line is $m = \frac {C_y - A_y}{C_x - A_x} = \frac {1-(-1)}{4-1} = \frac 23$ so the equation is $(y-A_y) = m*(x-A_x)$ so $y-(-1) = \frac 23(x-1)$ or $y = \frac 23 x -\frac 53$ or as you put it $2x - 3y = 5$.

2) What value of $K$ satisfies $2*\frac K2 - 3*\frac K3 = 5$ ?

$K - K = 5$

$0 = 5$

There is no value of $K$ that satisfies.

So the answer to "For what value of K are the points A(1, -1), B(K/2, K/3), and C(4, 1) collinear?" is...

"None".

Which is a perfectly acceptable and valid answer. Just because a textbook asks a question doesn't mean that there is an valid answer. But in this case "none" is a valid answer. Just because a textbook asks "when does this occur" doesn't mean the answer isn't "never".

Answer 2

If can you can proof that no real number $K$ will lead to a point $B$ which is on the line through $A$ and $C$, that is fine.

It might be an error in the book or for educational purposes.

My attempt:

The line through $A$ and $C$ is $$ (1-\lambda) A + \lambda C \quad (\lambda \in \mathbb{R}) $$ We equate with point $B$ and get $$ (1 - \lambda) (1, -1) + \lambda (4,1) = (K/2, K/3) $$ which gives the system $$ (1 - \lambda) + 4 \lambda = K/2 \\ -(1-\lambda) + \lambda = K/3 $$ which is equivalent to the inhomogeneous system in unknowns $\lambda$ and $K$: $$ \left[ \begin{array}{rr|r} 3 & -1/2 & -1 \\ 2 & -1/3 & 1 \end{array} \right] \to \left[ \begin{array}{rr|r} 1 & -1/6 & -2 \\ 2 & -1/3 & 1 \end{array} \right] \to \left[ \begin{array}{rr|r} 1 & -1/6 & -2 \\ 0 & 0 & 5 \end{array} \right] $$ We subtract the second row from the first row, then subtract two times the first row from the second row. The resulting last row is inconsistent ($0\cdot \lambda + 0 \cdot K = 5$), so there is no solution.

Answer 3

The direction vector of the straight line $K(1/2,1/3)$ is obviously parallel to the direction vector $C-A=(3,2)$ of the straight line through $A$ and $C$.