How do I find the gradient of BC? Two vertices of rectangle ABCD are A(3,-5), B(6,-3)

Kulla_9289

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Two vertices of a rectangle ABCD are A(3,-5) and B(6,-3).
a) Find the gradient of CD.
My working: C is (6-5) and D is (3,-3). The gradient is -2/3.
b) Find the gradient of BC. I am not sure about this.
 
Two vertices of a rectangle ABCD are A(3,-5) and B(6,-3).
a) Find the gradient of CD.
My working: C is (6-5) and D is (3,-3). The gradient is -2/3.
b) Find the gradient of BC. I am not sure about this.
Where did you get the coordinates for C & D???

Were they given in the question (you did not show this) or have you just 'guessed' them?

However, it looks to me like you have just assumed these were the coordinates of
C & D to make the 'nice' rectangle ACBD but plane figures usually have their vertices (corners) named in alphabetical order in either a clockwise or an anti-clockwise direction!

The gradient of CD (in your construction) is, indeed
\(\displaystyle -\frac{2}{3}\) but BC then becomes a vertical line and the gradient of that is not defined (would involve division by zero!) which is why you are struggling with it!


I suspect you have not fully understood the question(s) as you have presented them.

It is more likely that the rectangle you are supposed to consider would be like
ABC'D' (in my picture, qv) where C' & D' could lie anywhere along the lines they are sitting on.

The gradient of
CD will then be the same as AB (because they are parallel) and BC will be at right angles to CD so you should now be able to calculate the correct answers.

If not, come back and show us what you have done and further advice will be offered if necessary.

Please now show us your attempts in light of what I have told you...


Gradients.png

PS: This is not Intermediate/Advanced Algebra. It should more properly have been posted in the Geometry & Trig section of the forum.
 
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BC is perpendicular to AB. Do you know how to find a perpendicular slope?
Not in this OP's world!
S/he has assumed (incorrect) coordinate values for points
C & D making BC vertical (see Post #2 & diagram therein).
(S/he appears to know how to find the gradient of
AB so I was expecting him/her to come back & ask if s/he didn't know how the gradients of perpendicular lines are related.)
 
Of course I do. It's the negative reciprocal. @The Highlander, how do you know the rectangle should be like that; the way you have drawn it?
For the reasons outlined in Post #2 (qv). Have you read it????
If you put
C & D where you did then BC has no gradient! (And AB should be a side of the rectangle; it would be wrong to assume it was a diagonal.)
Since you have confirmed your knowledge of the gradients of perpendicular lines then you should require no further help.
I made no assumptions about what you did or didn't know (but the mistake you made led me to suspect that you might be a 'novice' in this area) which is why I just set you off on the right track and left it to you to figure out what was required to get the right answers (or ask for further help).
I'm not about to spend anymore time explaining what is perfectly obvious (from what I already took the trouble to provide).
If you want to show us your 'new' answers to parts
a) & b) then I'm sure someone will be happy to confirm them (or offer further advice if you're still not getting them right); otherwise...
 
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I'm not seeing where it was stated that the rectangle is necessarily orthogonal to the axes...
 
I'm not seeing where it was stated that the rectangle is necessarily orthogonal to the axes...
Very little was "stated" in the original question (as it was presented) but it is clear (IMNSHO) that this question is intended to test:-

a) the student's knowledge of how to calculate the gradient of a line when two points on it are given (and that the parallel sides of a rectangle share the same gradient)

and

b) the student's knowledge of the relationship between the gradients of perpendicular lines.
 
I'm not seeing where it was stated that the rectangle is necessarily orthogonal to the axes...
Furthermore, if I understand the statement you make correctly, then it's true that it's not "stated" anywhere "that the rectangle is necessarily orthogonal to the axes" and because that wasn't stated that's another good reason why the rectangle isn't orthogonal to the axes!
(But I suspect you were trying to say something other than what you did.)
 
I'm not seeing where it was stated that the rectangle is necessarily orthogonal to the axes...
This does appear to be what the OP wrongly assumed:
Two vertices of a rectangle ABCD are A(3,-5) and B(6,-3).
a) Find the gradient of CD.
My working: C is (6,-5) and D is (3,-3). The gradient is -2/3.
b) Find the gradient of BC. I am not sure about this.
The student has likely seen other problems in which this is appropriate reasoning, where the sides are said to be horizontal and vertical, and it is the diagonal that is given. So this is probably a case of not seeing what is actually stated, but what one is used to. It looks like a familiar problem, so let's do it that way.

This also requires taking "two vertices" to mean that they are not necessarily consecutive (and missing the convention for naming polygons consecutively); then it's at least the easiest possibility, and one might not notice that the problem asks about definite facts, not just possibilities. So it's somewhat understandable, though wrong on several counts.

Of course, even interpreting the problem correctly, we have no idea where C and D actually are, which might lead a student away from that interpretation, which seems to have too much uncertainty; in fact, the actual questions asked are just about all that can be asked.

All of this takes some experience to recognize. A student who is accustomed to fairly basic problems will easily miss the subtle implications that lead to the right interpretation.

Very little was "stated" in the original question (as it was presented) but it is clear (IMNSHO) that this question is intended to test:-

a) the student's knowledge of how to calculate the gradient of a line when two points on it are given (and that the parallel sides of a rectangle share the same gradient)


and

b) the student's knowledge of the relationship between the gradients of perpendicular lines.
Interestingly, although you are probably right about the goal, all sorts of subtleties in other areas get in the way of seeing it.
 
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