Implications of zero in the denominator when solving equations

[JC2000]

My book states the following :

${\frac{(y_1)-(y_2)}{(x_1)-(x_2)}}*\frac{(3+1)}{(0-0)} = -1$ ...(1)
$\implies y_1 - y_2 = 0$

(A) Is this a valid deduction?

Context :

The Problem :If $A(0,-1)$ and $B(0,3)$ are two opposite vertices of a square, then find the other two vertices.

Solution :

Let D and B be $(x_1,y_1)$ and $(x_2,y_2)$ respectively. Using A and B we can find the mid point of the diagonals M =(0,1).
Since M is also the mid point for BD we get : $x_1 + x_2 = 0$ and $y_1 + y_2 =2$
Also, (Slope of BD)*(Slope of AC) = -1 This is how statement (1) is arrived at.

(B) if this is an invalid attempt to solve the problem, how else could this be solved ( I realize that the problem with the given values can be solved by plotting the points and hence I am wondering if there is another general method to solve such a problem)?

 

[Mark44]

My book states the following :

${\frac{(y_1)-(y_2)}{(x_1)-(x_2)}}*\frac{(3+1)}{(0-0)} = -1$ ...(1)
$\implies y_1 - y_2 = 0$

Really? Your book has this?
It makes no sense to multiply something by $\frac 4 0$ and get -1 as the product.

(A) Is this a valid deduction?

Context :

The Problem :If $A(0,-1)$ and $B(0,3)$ are two opposite vertices of a square, then find the other two vertices.

Solution :

Let D and B be $(x_1,y_1)$ and $(x_2,y_2)$ respectively. Using A and B we can find the mid point of the diagonals M =(0,1).

This isn't consistent with what you wrote earlier, with A(0, -1) and B(0, 3). The other two vertices of the square would be C and D.

Since M is also the mid point for BD we get : $x_1 + x_2 = 0$ and $y_1 + y_2 =2$
Also, (Slope of BD)*(Slope of AC) = -1 This is how statement (1) is arrived at.

Based on what you wrote as the given information, this is all very confused. The diagonals would be AB (a vertical line segment) and CD (a horizontal line segment). It's still not legitimate to write an equation that multiplies 0 and something undefined to get -1.

(B) if this is an invalid attempt to solve the problem, how else could this be solved ( I realize that the problem with the given values can be solved by plotting the points and hence I am wondering if there is another general method to solve such a problem)?

The two given points are on the y-axis, and are 4 units apart. The other two vertices must be on a horizontal line, also 4 units apart. It should be fairly simple to figure out the coordinates of C and D.

 

[JC2000]

 

[Mark44]

That's terrible! From their drawing, one could immediately see that A and C are on the line x = 0 (the y-axis), which means that the slope of AC is undefined.

What's the name of this book and who is the author?

 

[JC2000]

Yes, I realized my error. I meant A and C were the known coordinates.
The book is actually a question bank, seems the publishers haven't mentioned the author's name.
Thank you for your detailed response.

 

[nuuskur]

The implication is true, indeed, but that doesn't make the conclusion true. The diagonals are perpendicular if and only if the product of their slopes is $-1$. This need not be the case.

I only see the parameters specified, but no problem posed.
Edit: nevermind, the problem is to figure out how to compute the midpoint coordinates.

The $0$ in the denominator phenomenon is an indication that something is wrong with the assumptions - in this case, the opposite corners $A$ and $C$ have the same $x$ coordinate, which is impossible.

 

[Mark44]

The implication is true, indeed, but that doesn't make the conclusion true. The diagonals are perpendicular if and only if the product of their slopes is −1. This need not be the case.

I think I understand what you meant, but that's different from what you wrote. I believe you mean that this statement -- The diagonals are perpendicular if and only if the product of their slopes is −1. -- is not necessarily true. It's certainly false if one of the diagonals of the square is vertical.

The 0 in the denominator phenomenon is an indication that something is wrong with the assumptions - in this case, the opposite corners A and C have the same x coordinate, which is impossible.

No, not impossible if square is rotated by an angle of 45° relative to the x-axis.