Line distance between points (p,q) and (q,p)
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I'm trying to find the distance between the points (p,q) and (q,p). As far as I can tell, my steps are correct, but I'm getting the answer
$sqrt{2q^2 + 2p^2}$ but the textbook gives the answer $sqrt{2(p-q)^2}$. Can anyone check my steps please, and tell me what, if anything I'm doing wrong:
begin{split}
x_2 &= q\
x_1 &= p\
y_2 &= p\
y_1 &= q\
\
x_2 - x_1 &= q - p\
y_2 - y_1 &= p - q\
\
text{ distance }&= sqrt{(x_2 - x_1)^2} + sqrt{(y_2 - y_1)^2}\
\
text{ (plugging} &text{ in values)}\
\
text{ distance } &=sqrt{(q-p)^2} + sqrt{(p-q)^2}\
\
text{($-p$ and $-q$ squared turn to $p^2$ } &text{ and $q^2$, removing negative signs:)}\
\
&=sqrt{q^2} + sqrt{p^2} + sqrt{p^2} + sqrt{q^2}\
\
&=sqrt{2q^2} + sqrt{2p^2}
end{split}
The textbook answer however is $sqrt{2(p-q)^2}$. Re-reading it, it did say to assume $p > q > 0$. Would that have changed the values / results?
Thanks and sorry about the formatting - first time poster.
analytic-geometry coordinate-systems
asked Dec 3 at 17:56
Click Upvote
1136
add a comment |
up vote
-1
down vote
favorite
I'm trying to find the distance between the points (p,q) and (q,p). As far as I can tell, my steps are correct, but I'm getting the answer
$sqrt{2q^2 + 2p^2}$ but the textbook gives the answer $sqrt{2(p-q)^2}$. Can anyone check my steps please, and tell me what, if anything I'm doing wrong:
begin{split}
x_2 &= q\
x_1 &= p\
y_2 &= p\
y_1 &= q\
\
x_2 - x_1 &= q - p\
y_2 - y_1 &= p - q\
\
text{ distance }&= sqrt{(x_2 - x_1)^2} + sqrt{(y_2 - y_1)^2}\
\
text{ (plugging} &text{ in values)}\
\
text{ distance } &=sqrt{(q-p)^2} + sqrt{(p-q)^2}\
\
text{($-p$ and $-q$ squared turn to $p^2$ } &text{ and $q^2$, removing negative signs:)}\
\
&=sqrt{q^2} + sqrt{p^2} + sqrt{p^2} + sqrt{q^2}\
\
&=sqrt{2q^2} + sqrt{2p^2}
end{split}
The textbook answer however is $sqrt{2(p-q)^2}$. Re-reading it, it did say to assume $p > q > 0$. Would that have changed the values / results?
Thanks and sorry about the formatting - first time poster.
analytic-geometry coordinate-systems
asked Dec 3 at 17:56
Click Upvote
1136
2
You can't manipulate square roots that way. $sqrt {A+B}neq sqrt A +sqrt B$ unless at least one of $A,B$ is $0$.
– lulu
Dec 3 at 17:59
Thanks, that was just my attempt at having the square root look like it dragged across both numbers. :P
– Click Upvote
Dec 3 at 18:24
1
No...you systematically wrote things like $sqrt {(p-q)^2}+sqrt {(q-p)^2}$ where you should have written $sqrt {(p-q)^2+(q-p)^2}$. Similarly, you appear to believe that $(p-q)^2 =p^2+q^2$ which is false.
– lulu
Dec 3 at 18:49
Thanks for clearing that up
– Click Upvote
Dec 3 at 18:54
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I'm trying to find the distance between the points (p,q) and (q,p). As far as I can tell, my steps are correct, but I'm getting the answer
$sqrt{2q^2 + 2p^2}$ but the textbook gives the answer $sqrt{2(p-q)^2}$. Can anyone check my steps please, and tell me what, if anything I'm doing wrong:
begin{split}
x_2 &= q\
x_1 &= p\
y_2 &= p\
y_1 &= q\
\
x_2 - x_1 &= q - p\
y_2 - y_1 &= p - q\
\
text{ distance }&= sqrt{(x_2 - x_1)^2} + sqrt{(y_2 - y_1)^2}\
\
text{ (plugging} &text{ in values)}\
\
text{ distance } &=sqrt{(q-p)^2} + sqrt{(p-q)^2}\
\
text{($-p$ and $-q$ squared turn to $p^2$ } &text{ and $q^2$, removing negative signs:)}\
\
&=sqrt{q^2} + sqrt{p^2} + sqrt{p^2} + sqrt{q^2}\
\
&=sqrt{2q^2} + sqrt{2p^2}
end{split}
The textbook answer however is $sqrt{2(p-q)^2}$. Re-reading it, it did say to assume $p > q > 0$. Would that have changed the values / results?
Thanks and sorry about the formatting - first time poster.
analytic-geometry coordinate-systems
asked Dec 3 at 17:56
Click Upvote
1136
I'm trying to find the distance between the points (p,q) and (q,p). As far as I can tell, my steps are correct, but I'm getting the answer
$sqrt{2q^2 + 2p^2}$ but the textbook gives the answer $sqrt{2(p-q)^2}$. Can anyone check my steps please, and tell me what, if anything I'm doing wrong:
begin{split}
x_2 &= q\
x_1 &= p\
y_2 &= p\
y_1 &= q\
\
x_2 - x_1 &= q - p\
y_2 - y_1 &= p - q\
\
text{ distance }&= sqrt{(x_2 - x_1)^2} + sqrt{(y_2 - y_1)^2}\
\
text{ (plugging} &text{ in values)}\
\
text{ distance } &=sqrt{(q-p)^2} + sqrt{(p-q)^2}\
\
text{($-p$ and $-q$ squared turn to $p^2$ } &text{ and $q^2$, removing negative signs:)}\
\
&=sqrt{q^2} + sqrt{p^2} + sqrt{p^2} + sqrt{q^2}\
\
&=sqrt{2q^2} + sqrt{2p^2}
end{split}
The textbook answer however is $sqrt{2(p-q)^2}$. Re-reading it, it did say to assume $p > q > 0$. Would that have changed the values / results?
Thanks and sorry about the formatting - first time poster.
analytic-geometry coordinate-systems
analytic-geometry coordinate-systems
asked Dec 3 at 17:56
Click Upvote
1136
asked Dec 3 at 17:56
Click Upvote
1136
edited Dec 4 at 22:21
asked Dec 3 at 17:56
Click Upvote
1136
asked Dec 3 at 17:56
Click Upvote
1136
asked Dec 3 at 17:56
Click Upvote
1136
1136
2
You can't manipulate square roots that way. $sqrt {A+B}neq sqrt A +sqrt B$ unless at least one of $A,B$ is $0$.
– lulu
Dec 3 at 17:59
Thanks, that was just my attempt at having the square root look like it dragged across both numbers. :P
– Click Upvote
Dec 3 at 18:24
1
No...you systematically wrote things like $sqrt {(p-q)^2}+sqrt {(q-p)^2}$ where you should have written $sqrt {(p-q)^2+(q-p)^2}$. Similarly, you appear to believe that $(p-q)^2 =p^2+q^2$ which is false.
– lulu
Dec 3 at 18:49
Thanks for clearing that up
– Click Upvote
Dec 3 at 18:54
add a comment |
2
You can't manipulate square roots that way. $sqrt {A+B}neq sqrt A +sqrt B$ unless at least one of $A,B$ is $0$.
– lulu
Dec 3 at 17:59
Thanks, that was just my attempt at having the square root look like it dragged across both numbers. :P
– Click Upvote
Dec 3 at 18:24
1
No...you systematically wrote things like $sqrt {(p-q)^2}+sqrt {(q-p)^2}$ where you should have written $sqrt {(p-q)^2+(q-p)^2}$. Similarly, you appear to believe that $(p-q)^2 =p^2+q^2$ which is false.
– lulu
Dec 3 at 18:49
Thanks for clearing that up
– Click Upvote
Dec 3 at 18:54
2
2
You can't manipulate square roots that way. $sqrt {A+B}neq sqrt A +sqrt B$ unless at least one of $A,B$ is $0$.
– lulu
Dec 3 at 17:59
You can't manipulate square roots that way. $sqrt {A+B}neq sqrt A +sqrt B$ unless at least one of $A,B$ is $0$.
– lulu
Dec 3 at 17:59
Thanks, that was just my attempt at having the square root look like it dragged across both numbers. :P
– Click Upvote
Dec 3 at 18:24
Thanks, that was just my attempt at having the square root look like it dragged across both numbers. :P
– Click Upvote
Dec 3 at 18:24
1
1
No...you systematically wrote things like $sqrt {(p-q)^2}+sqrt {(q-p)^2}$ where you should have written $sqrt {(p-q)^2+(q-p)^2}$. Similarly, you appear to believe that $(p-q)^2 =p^2+q^2$ which is false.
– lulu
Dec 3 at 18:49
No...you systematically wrote things like $sqrt {(p-q)^2}+sqrt {(q-p)^2}$ where you should have written $sqrt {(p-q)^2+(q-p)^2}$. Similarly, you appear to believe that $(p-q)^2 =p^2+q^2$ which is false.
– lulu
Dec 3 at 18:49
Thanks for clearing that up
– Click Upvote
Dec 3 at 18:54
Thanks for clearing that up
– Click Upvote
Dec 3 at 18:54
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
The square root doesn't distribute. That is to say,
$$sqrt{a + b} neq sqrt{a} + sqrt{b}$$
in general. (There are specific examples and cases where it does equal and looks like it distributes, but it doesn't always happen. For example, above, consider $a=4,b=9$.)
Therein lies one of your problems.
Also, we note, once you get to the step where
$$d = sqrt{(q-p)^2 + √(p-q)^2}$$
notice that $(p-q)^2 = (q-p)^2$. If you have trouble convincing yourself of this, remember you can factor out a $-1$ from one of them and thus:
$$(q-p)=-1(p-q)$$
implying
$$(q-p)^2 = (q-p)(q-p) = (-1)(p-q)(-1)(p-q) = (-1)^2(p-q)^2 = (p-q)^2$$
This is probably the most confusing step of the textbook's solution. From there it should be clear how to proceed.
answered Dec 3 at 18:03
Eevee Trainer
2,214220
I know the square root doesn't distribute - I couldn't figure out how to format it so to be clear, I added it to both numbers. But, thanks a bunch for explaining that(q-p)^2 = (p-q)^2. Do you happen to know if there are any khan academy videos or such explaining this more so I can look more into this? Thanks.
– Click Upvote
Dec 3 at 18:17
I think(q-p)^2evaluates toq^2 + p^2and(p-q)^2evaluates to the same thing, i.ep^2 + q^2, and that's why they are both equal and can be factored out. Is my understanding correct?
– Click Upvote
Dec 3 at 18:23
No. Just like the square root, exponentiation doesn't distribute. Remember: $$(q-p)^2 = (q-p)(q-p)$$ Try FOILing that out, and see what you get.
– Eevee Trainer
Dec 3 at 18:26
I'm just trying to understand why(q-p)^2 = (p-q)^2. My attempt was that they evaluate to the same thing if you open the brackets. Is there another indicator to recognize situations like this? Also, the answer would be valid either as2(p-q)^2or2(q-p)^2since they are equal - right?
– Click Upvote
Dec 3 at 18:32
1
An alternative way to see that $(q-p)^2 = (p-q)^2$, aside from what I showed, would be to FOIL out each expression, since $$(q-p)^2 = (q-p)(q-p) ;;; text{and} ;;; (p-q)^2 = (p-q)(p-q)$$ I wouldn't know about any sort of "indicator" for situations like this, you just learn to recognize them with experience. And yes, you could use $2(p-q)^2$ or $2(q-p)^2$ since they're equal; which of the two you choose is purely up to you.
– Eevee Trainer
Dec 3 at 18:35
add a comment |
up vote
1
down vote
The distance is
$$
d = sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
$$
If you replace your points
$$
d = sqrt{(p - q)^2 + (q - p)^2} = sqrt{(p - q)^2 + (p - q)^2} = sqrt{2(p - q)^2} = sqrt{2} |p - q|
$$
answered Dec 3 at 17:58
caverac
12.5k21027
Thank you for the steps, this helped me see the solution, but the other answer's explanation helped me understand why, so i had to accept that.
– Click Upvote
Dec 3 at 18:18
1
@ClickUpvote I’m glad you found the solution you’re looking for
– caverac
Dec 3 at 18:24
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The square root doesn't distribute. That is to say,
$$sqrt{a + b} neq sqrt{a} + sqrt{b}$$
in general. (There are specific examples and cases where it does equal and looks like it distributes, but it doesn't always happen. For example, above, consider $a=4,b=9$.)
Therein lies one of your problems.
Also, we note, once you get to the step where
$$d = sqrt{(q-p)^2 + √(p-q)^2}$$
notice that $(p-q)^2 = (q-p)^2$. If you have trouble convincing yourself of this, remember you can factor out a $-1$ from one of them and thus:
$$(q-p)=-1(p-q)$$
implying
$$(q-p)^2 = (q-p)(q-p) = (-1)(p-q)(-1)(p-q) = (-1)^2(p-q)^2 = (p-q)^2$$
This is probably the most confusing step of the textbook's solution. From there it should be clear how to proceed.
answered Dec 3 at 18:03
Eevee Trainer
2,214220
I know the square root doesn't distribute - I couldn't figure out how to format it so to be clear, I added it to both numbers. But, thanks a bunch for explaining that(q-p)^2 = (p-q)^2. Do you happen to know if there are any khan academy videos or such explaining this more so I can look more into this? Thanks.
– Click Upvote
Dec 3 at 18:17
I think(q-p)^2evaluates toq^2 + p^2and(p-q)^2evaluates to the same thing, i.ep^2 + q^2, and that's why they are both equal and can be factored out. Is my understanding correct?
– Click Upvote
Dec 3 at 18:23
No. Just like the square root, exponentiation doesn't distribute. Remember: $$(q-p)^2 = (q-p)(q-p)$$ Try FOILing that out, and see what you get.
– Eevee Trainer
Dec 3 at 18:26
I'm just trying to understand why(q-p)^2 = (p-q)^2. My attempt was that they evaluate to the same thing if you open the brackets. Is there another indicator to recognize situations like this? Also, the answer would be valid either as2(p-q)^2or2(q-p)^2since they are equal - right?
– Click Upvote
Dec 3 at 18:32
1
An alternative way to see that $(q-p)^2 = (p-q)^2$, aside from what I showed, would be to FOIL out each expression, since $$(q-p)^2 = (q-p)(q-p) ;;; text{and} ;;; (p-q)^2 = (p-q)(p-q)$$ I wouldn't know about any sort of "indicator" for situations like this, you just learn to recognize them with experience. And yes, you could use $2(p-q)^2$ or $2(q-p)^2$ since they're equal; which of the two you choose is purely up to you.
– Eevee Trainer
Dec 3 at 18:35
add a comment |
up vote
1
down vote
accepted
The square root doesn't distribute. That is to say,
$$sqrt{a + b} neq sqrt{a} + sqrt{b}$$
in general. (There are specific examples and cases where it does equal and looks like it distributes, but it doesn't always happen. For example, above, consider $a=4,b=9$.)
Therein lies one of your problems.
Also, we note, once you get to the step where
$$d = sqrt{(q-p)^2 + √(p-q)^2}$$
notice that $(p-q)^2 = (q-p)^2$. If you have trouble convincing yourself of this, remember you can factor out a $-1$ from one of them and thus:
$$(q-p)=-1(p-q)$$
implying
$$(q-p)^2 = (q-p)(q-p) = (-1)(p-q)(-1)(p-q) = (-1)^2(p-q)^2 = (p-q)^2$$
This is probably the most confusing step of the textbook's solution. From there it should be clear how to proceed.
answered Dec 3 at 18:03
Eevee Trainer
2,214220
I know the square root doesn't distribute - I couldn't figure out how to format it so to be clear, I added it to both numbers. But, thanks a bunch for explaining that(q-p)^2 = (p-q)^2. Do you happen to know if there are any khan academy videos or such explaining this more so I can look more into this? Thanks.
– Click Upvote
Dec 3 at 18:17
I think(q-p)^2evaluates toq^2 + p^2and(p-q)^2evaluates to the same thing, i.ep^2 + q^2, and that's why they are both equal and can be factored out. Is my understanding correct?
– Click Upvote
Dec 3 at 18:23
No. Just like the square root, exponentiation doesn't distribute. Remember: $$(q-p)^2 = (q-p)(q-p)$$ Try FOILing that out, and see what you get.
– Eevee Trainer
Dec 3 at 18:26
I'm just trying to understand why(q-p)^2 = (p-q)^2. My attempt was that they evaluate to the same thing if you open the brackets. Is there another indicator to recognize situations like this? Also, the answer would be valid either as2(p-q)^2or2(q-p)^2since they are equal - right?
– Click Upvote
Dec 3 at 18:32
1
An alternative way to see that $(q-p)^2 = (p-q)^2$, aside from what I showed, would be to FOIL out each expression, since $$(q-p)^2 = (q-p)(q-p) ;;; text{and} ;;; (p-q)^2 = (p-q)(p-q)$$ I wouldn't know about any sort of "indicator" for situations like this, you just learn to recognize them with experience. And yes, you could use $2(p-q)^2$ or $2(q-p)^2$ since they're equal; which of the two you choose is purely up to you.
– Eevee Trainer
Dec 3 at 18:35
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The square root doesn't distribute. That is to say,
$$sqrt{a + b} neq sqrt{a} + sqrt{b}$$
in general. (There are specific examples and cases where it does equal and looks like it distributes, but it doesn't always happen. For example, above, consider $a=4,b=9$.)
Therein lies one of your problems.
Also, we note, once you get to the step where
$$d = sqrt{(q-p)^2 + √(p-q)^2}$$
notice that $(p-q)^2 = (q-p)^2$. If you have trouble convincing yourself of this, remember you can factor out a $-1$ from one of them and thus:
$$(q-p)=-1(p-q)$$
implying
$$(q-p)^2 = (q-p)(q-p) = (-1)(p-q)(-1)(p-q) = (-1)^2(p-q)^2 = (p-q)^2$$
This is probably the most confusing step of the textbook's solution. From there it should be clear how to proceed.
answered Dec 3 at 18:03
Eevee Trainer
2,214220
The square root doesn't distribute. That is to say,
$$sqrt{a + b} neq sqrt{a} + sqrt{b}$$
in general. (There are specific examples and cases where it does equal and looks like it distributes, but it doesn't always happen. For example, above, consider $a=4,b=9$.)
Therein lies one of your problems.
Also, we note, once you get to the step where
$$d = sqrt{(q-p)^2 + √(p-q)^2}$$
notice that $(p-q)^2 = (q-p)^2$. If you have trouble convincing yourself of this, remember you can factor out a $-1$ from one of them and thus:
$$(q-p)=-1(p-q)$$
implying
$$(q-p)^2 = (q-p)(q-p) = (-1)(p-q)(-1)(p-q) = (-1)^2(p-q)^2 = (p-q)^2$$
This is probably the most confusing step of the textbook's solution. From there it should be clear how to proceed.
answered Dec 3 at 18:03
Eevee Trainer
2,214220
answered Dec 3 at 18:03
Eevee Trainer
2,214220
answered Dec 3 at 18:03
Eevee Trainer
2,214220
answered Dec 3 at 18:03
Eevee Trainer
2,214220
2,214220
I know the square root doesn't distribute - I couldn't figure out how to format it so to be clear, I added it to both numbers. But, thanks a bunch for explaining that(q-p)^2 = (p-q)^2. Do you happen to know if there are any khan academy videos or such explaining this more so I can look more into this? Thanks.
– Click Upvote
Dec 3 at 18:17
I think(q-p)^2evaluates toq^2 + p^2and(p-q)^2evaluates to the same thing, i.ep^2 + q^2, and that's why they are both equal and can be factored out. Is my understanding correct?
– Click Upvote
Dec 3 at 18:23
No. Just like the square root, exponentiation doesn't distribute. Remember: $$(q-p)^2 = (q-p)(q-p)$$ Try FOILing that out, and see what you get.
– Eevee Trainer
Dec 3 at 18:26
I'm just trying to understand why(q-p)^2 = (p-q)^2. My attempt was that they evaluate to the same thing if you open the brackets. Is there another indicator to recognize situations like this? Also, the answer would be valid either as2(p-q)^2or2(q-p)^2since they are equal - right?
– Click Upvote
Dec 3 at 18:32
1
An alternative way to see that $(q-p)^2 = (p-q)^2$, aside from what I showed, would be to FOIL out each expression, since $$(q-p)^2 = (q-p)(q-p) ;;; text{and} ;;; (p-q)^2 = (p-q)(p-q)$$ I wouldn't know about any sort of "indicator" for situations like this, you just learn to recognize them with experience. And yes, you could use $2(p-q)^2$ or $2(q-p)^2$ since they're equal; which of the two you choose is purely up to you.
– Eevee Trainer
Dec 3 at 18:35
add a comment |
I know the square root doesn't distribute - I couldn't figure out how to format it so to be clear, I added it to both numbers. But, thanks a bunch for explaining that(q-p)^2 = (p-q)^2. Do you happen to know if there are any khan academy videos or such explaining this more so I can look more into this? Thanks.
– Click Upvote
Dec 3 at 18:17
I think(q-p)^2evaluates toq^2 + p^2and(p-q)^2evaluates to the same thing, i.ep^2 + q^2, and that's why they are both equal and can be factored out. Is my understanding correct?
– Click Upvote
Dec 3 at 18:23
No. Just like the square root, exponentiation doesn't distribute. Remember: $$(q-p)^2 = (q-p)(q-p)$$ Try FOILing that out, and see what you get.
– Eevee Trainer
Dec 3 at 18:26
I'm just trying to understand why(q-p)^2 = (p-q)^2. My attempt was that they evaluate to the same thing if you open the brackets. Is there another indicator to recognize situations like this? Also, the answer would be valid either as2(p-q)^2or2(q-p)^2since they are equal - right?
– Click Upvote
Dec 3 at 18:32
1
An alternative way to see that $(q-p)^2 = (p-q)^2$, aside from what I showed, would be to FOIL out each expression, since $$(q-p)^2 = (q-p)(q-p) ;;; text{and} ;;; (p-q)^2 = (p-q)(p-q)$$ I wouldn't know about any sort of "indicator" for situations like this, you just learn to recognize them with experience. And yes, you could use $2(p-q)^2$ or $2(q-p)^2$ since they're equal; which of the two you choose is purely up to you.
– Eevee Trainer
Dec 3 at 18:35
I know the square root doesn't distribute - I couldn't figure out how to format it so to be clear, I added it to both numbers. But, thanks a bunch for explaining that
(q-p)^2 = (p-q)^2 . Do you happen to know if there are any khan academy videos or such explaining this more so I can look more into this? Thanks.– Click Upvote
Dec 3 at 18:17
I know the square root doesn't distribute - I couldn't figure out how to format it so to be clear, I added it to both numbers. But, thanks a bunch for explaining that
(q-p)^2 = (p-q)^2 . Do you happen to know if there are any khan academy videos or such explaining this more so I can look more into this? Thanks.– Click Upvote
Dec 3 at 18:17
I think
(q-p)^2 evaluates to q^2 + p^2 and (p-q)^2 evaluates to the same thing, i.e p^2 + q^2, and that's why they are both equal and can be factored out. Is my understanding correct?– Click Upvote
Dec 3 at 18:23
I think
(q-p)^2 evaluates to q^2 + p^2 and (p-q)^2 evaluates to the same thing, i.e p^2 + q^2, and that's why they are both equal and can be factored out. Is my understanding correct?– Click Upvote
Dec 3 at 18:23
No. Just like the square root, exponentiation doesn't distribute. Remember: $$(q-p)^2 = (q-p)(q-p)$$ Try FOILing that out, and see what you get.
– Eevee Trainer
Dec 3 at 18:26
No. Just like the square root, exponentiation doesn't distribute. Remember: $$(q-p)^2 = (q-p)(q-p)$$ Try FOILing that out, and see what you get.
– Eevee Trainer
Dec 3 at 18:26
I'm just trying to understand why
(q-p)^2 = (p-q)^2. My attempt was that they evaluate to the same thing if you open the brackets. Is there another indicator to recognize situations like this? Also, the answer would be valid either as 2(p-q)^2 or 2(q-p)^2 since they are equal - right?– Click Upvote
Dec 3 at 18:32
I'm just trying to understand why
(q-p)^2 = (p-q)^2. My attempt was that they evaluate to the same thing if you open the brackets. Is there another indicator to recognize situations like this? Also, the answer would be valid either as 2(p-q)^2 or 2(q-p)^2 since they are equal - right?– Click Upvote
Dec 3 at 18:32
1
1
An alternative way to see that $(q-p)^2 = (p-q)^2$, aside from what I showed, would be to FOIL out each expression, since $$(q-p)^2 = (q-p)(q-p) ;;; text{and} ;;; (p-q)^2 = (p-q)(p-q)$$ I wouldn't know about any sort of "indicator" for situations like this, you just learn to recognize them with experience. And yes, you could use $2(p-q)^2$ or $2(q-p)^2$ since they're equal; which of the two you choose is purely up to you.
– Eevee Trainer
Dec 3 at 18:35
An alternative way to see that $(q-p)^2 = (p-q)^2$, aside from what I showed, would be to FOIL out each expression, since $$(q-p)^2 = (q-p)(q-p) ;;; text{and} ;;; (p-q)^2 = (p-q)(p-q)$$ I wouldn't know about any sort of "indicator" for situations like this, you just learn to recognize them with experience. And yes, you could use $2(p-q)^2$ or $2(q-p)^2$ since they're equal; which of the two you choose is purely up to you.
– Eevee Trainer
Dec 3 at 18:35
add a comment |
up vote
1
down vote
The distance is
$$
d = sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
$$
If you replace your points
$$
d = sqrt{(p - q)^2 + (q - p)^2} = sqrt{(p - q)^2 + (p - q)^2} = sqrt{2(p - q)^2} = sqrt{2} |p - q|
$$
answered Dec 3 at 17:58
caverac
12.5k21027
Thank you for the steps, this helped me see the solution, but the other answer's explanation helped me understand why, so i had to accept that.
– Click Upvote
Dec 3 at 18:18
1
@ClickUpvote I’m glad you found the solution you’re looking for
– caverac
Dec 3 at 18:24
add a comment |
up vote
1
down vote
The distance is
$$
d = sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
$$
If you replace your points
$$
d = sqrt{(p - q)^2 + (q - p)^2} = sqrt{(p - q)^2 + (p - q)^2} = sqrt{2(p - q)^2} = sqrt{2} |p - q|
$$
answered Dec 3 at 17:58
caverac
12.5k21027
Thank you for the steps, this helped me see the solution, but the other answer's explanation helped me understand why, so i had to accept that.
– Click Upvote
Dec 3 at 18:18
1
@ClickUpvote I’m glad you found the solution you’re looking for
– caverac
Dec 3 at 18:24
add a comment |
up vote
1
down vote
up vote
1
down vote
The distance is
$$
d = sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
$$
If you replace your points
$$
d = sqrt{(p - q)^2 + (q - p)^2} = sqrt{(p - q)^2 + (p - q)^2} = sqrt{2(p - q)^2} = sqrt{2} |p - q|
$$
answered Dec 3 at 17:58
caverac
12.5k21027
The distance is
$$
d = sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
$$
If you replace your points
$$
d = sqrt{(p - q)^2 + (q - p)^2} = sqrt{(p - q)^2 + (p - q)^2} = sqrt{2(p - q)^2} = sqrt{2} |p - q|
$$
answered Dec 3 at 17:58
caverac
12.5k21027
answered Dec 3 at 17:58
caverac
12.5k21027
answered Dec 3 at 17:58
caverac
12.5k21027
answered Dec 3 at 17:58
caverac
12.5k21027
12.5k21027
Thank you for the steps, this helped me see the solution, but the other answer's explanation helped me understand why, so i had to accept that.
– Click Upvote
Dec 3 at 18:18
1
@ClickUpvote I’m glad you found the solution you’re looking for
– caverac
Dec 3 at 18:24
add a comment |
Thank you for the steps, this helped me see the solution, but the other answer's explanation helped me understand why, so i had to accept that.
– Click Upvote
Dec 3 at 18:18
1
@ClickUpvote I’m glad you found the solution you’re looking for
– caverac
Dec 3 at 18:24
Thank you for the steps, this helped me see the solution, but the other answer's explanation helped me understand why, so i had to accept that.
– Click Upvote
Dec 3 at 18:18
Thank you for the steps, this helped me see the solution, but the other answer's explanation helped me understand why, so i had to accept that.
– Click Upvote
Dec 3 at 18:18
1
1
@ClickUpvote I’m glad you found the solution you’re looking for
– caverac
Dec 3 at 18:24
@ClickUpvote I’m glad you found the solution you’re looking for
– caverac
Dec 3 at 18:24
add a comment |
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2
You can't manipulate square roots that way. $sqrt {A+B}neq sqrt A +sqrt B$ unless at least one of $A,B$ is $0$.
– lulu
Dec 3 at 17:59
Thanks, that was just my attempt at having the square root look like it dragged across both numbers. :P
– Click Upvote
Dec 3 at 18:24
1
No...you systematically wrote things like $sqrt {(p-q)^2}+sqrt {(q-p)^2}$ where you should have written $sqrt {(p-q)^2+(q-p)^2}$. Similarly, you appear to believe that $(p-q)^2 =p^2+q^2$ which is false.
– lulu
Dec 3 at 18:49
Thanks for clearing that up
– Click Upvote
Dec 3 at 18:54