MHB Is there a way to view LaTex on a cell phone?

[mathdad]

Points P and Q are reflected in the line y = x to obtain points P' and Q', respectively. Does the distance from P to Q = the distance from P' to Q'?

What's the set up here?

 

[Evgeny.Makarov]

Any reflection across a line in the plane is an isometry (a distance-preserving transformation). In this special case it can be shown algebraically as follows. You know from another problem that the reflection of a point $(x,y)$ in the line $y=x$ has coordinates $(y,x)$. Let $P(x_1,y_1)$, $Q(x_2,y_2)$. Write the coordinates of $P'$ and $Q'$ and the distances $PQ$ and $P'Q'$.

 

[mathdad]

Any reflection across a line in the plane is an isometry (a distance-preserving transformation). In this special case it can be shown algebraically as follows. You know from another problem that the reflection of a point $(x,y)$ in the line $y=x$ has coordinates $(y,x)$. Let $P(x_1,y_1)$, $Q(x_2,y_2)$. Write the coordinates of $P'$ and $Q'$ and the distances $PQ$ and $P'Q'$.

Are you saying to use the distance formula for points?

 

[Evgeny.Makarov]

Are you saying to use the distance formula for points?

Yes. The fact that $PQ=P'Q'$ can also be shown geometrically.

 

[mathdad]

Yes. The fact that $PQ=P'Q'$ can also be shown geometrically.

How is this shown geometrically?

 

[Evgeny.Makarov]

Suppose that $QQ'>PP'$. Drop perpendiculars $PR$ and $P'R'$ on $QQ'$. Let $M$ and $N$ be the midpoints of $PP'$ and $QQ'$, respectively. Then $PRNM$ and $P'R'NM$ are equal rectangles. Therefore $PR=P'R'$ and $QR=QN-RN=Q'N-R'N=Q'R'$. So $\triangle PQR=\triangle P'Q'R'$ and $PQ=P'Q'$.

Alternatively, Wikipedia says that for the trapezoid $PP'Q'Q$ to be isosceles it is sufficient that the segment $MN$ that joins the midpoints of the parallel sides is perpendicular to them, which is the case here by definition of symmetry.

 

[mathdad]

Suppose that $QQ'>PP'$. Drop perpendiculars $PR$ and $P'R'$ on $QQ'$. Let $M$ and $N$ be the midpoints of $PP'$ and $QQ'$, respectively. Then $PRNM$ and $P'R'NM$ are equal rectangles. Therefore $PR=P'R'$ and $QR=QN-RN=Q'N-R'N=Q'R'$. So $\triangle PQR=\triangle P'Q'R'$ and $PQ=P'Q'$.

Alternatively, Wikipedia says that for the trapezoid $PP'Q'Q$ to be isosceles it is sufficient that the segment $MN$ that joins the midpoints of the parallel sides is perpendicular to them, which is the case here by definition of symmetry.

Your latex reply overlaps. I cannot read it.

 

[Evgeny.Makarov]

Did you notice any problem with displaying LaTeX on other MHB pages? Try disabling any JavaScript blocker such as NoScript. If this does not help, the staff would appreciate if you submit a report in the http://mathhelpboards.com/questions-comments-feedback-25/ with a screenshot and browser version.

 

[mathdad]

Did you notice any problem with displaying LaTeX on other MHB pages? Try disabling any JavaScript blocker such as NoScript. If this does not help, the staff would appreciate if you submit a report in the http://mathhelpboards.com/questions-comments-feedback-25/ with a screenshot and browser version.

I do not care about LaTex.

 

[MarkFL]

I do not care about LaTex.

The vast majority of math helpers here, and indeed on every other math help site I know of, are going to use $\LaTeX$ when responding to questions. Reading anything but the simplest of expressions formatted in plain text is a chore at best.

So, if you care about being able to read the help with which you will be provided, it would be in your best interest to take steps to ensure you can read it. It may be as simple as using a better browser. :D

 

[mathdad]

The vast majority of math helpers here, and indeed on every other math help site I know of, are going to use $\LaTeX$ when responding to questions. Reading anything but the simplest of expressions formatted in plain text is a chore at best.

So, if you care about being able to read the help with which you will be provided, it would be in your best interest to take steps to ensure you can read it. It may be as simple as using a better browser. :D

I do not have a computer or laptop. All my questions and replies are done via cell phone.

 

[MarkFL]

I do not have a computer or laptop. All my questions and replies are done via cell phone.

Do you not have any way to try other browsers?

 

[mathdad]

Do you not have any way to try other browsers?

No but I can see each reply given in LaTex form on my cell phone.