I have 2 lines, and I want to get the point where they intersect. I am looking to have a point created on the point of intersection. How can this be done?
22777
I need to put a point at the intersection of two edges. Below is a picture with red circles around where I need points to be placed.
81958
I have a self-intersecting line that I tried to split using the Split Features tool in the Advanced Digitizing Toolbar. The problem is that when I try to split the line at one point, the line splits at every point where it crosses over itself.  This line was originally a tracklog generated in WGS 84 using a Garmin GPS. I imported it into QGIS in GPX format, and saved as a shapefile.  The line is styled with a red marker at the first node.  Before splitting at the blue circle:   After splitting:    What's going wrong here?
116980
I need to fill the region determined by two (Gaussian) curves and the right red line (the region is brushed in green in the figure below). I also wonder what is the best way to draw vertical line which goes through the intersection of both graphs (the left red line on the figure).  Any pointers would be greatly appreciated.    My initial attemt is pasted below.  \documentclass{standalone} \usepackage{pgfplots}  \begin{document}  \pgfmathdeclarefunction{dnorm}{2}{%   \pgfmathparse{1/(#2*sqrt(2*pi))*exp(-((x-#1)^2)/(2*#2^2))}% }  \begin{tikzpicture} \begin{axis}[domain=0:12, samples=100, height=5cm, width=10cm] % Fill aread under the curves \addplot [fill=red!20, draw=none, domain=0:6] {dnorm(6.5,1.5)} \closedcycle; \addplot [fill=blue!20, draw=none, domain=6:10] {dnorm(4,1)} \closedcycle; % Draw curves \addplot [thick] {dnorm(4,1)}; \addplot [thick] {dnorm(6.5,1.5)}; \end{axis} \end{tikzpicture}  \end{document}   My solution  A picture is worth a thousand words. Code is pasted below.   \documentclass{article} \usepackage{tkz-fct} \usetikzlibrary{intersections}  \begin{document}  \tikzset{    name plot/.style={every path/.style={name path global=#1}} }  % Extract coordinates for point X \makeatletter \newcommand{\gettikzxy}[3]{%   \tikz@scan@one@point\pgfutil@firstofone#1\relax   \edef#2{\the\pgf@x}%   \edef#3{\the\pgf@y}% } \makeatother  % Dimlines \def\Dimline[#1][#2][#3][#4]{ \begin{scope}[thin, &gt;=stealth'] % redefine as flechas \draw let \p1=#1, \p2=#2, \n0={veclen(\x2-\x1,\y2-\y1)} in [|&lt;-&gt;|, decoration={markings,mark=at position .5 with {\node[#3] at (0,0) {#4};}, }, postaction=decorate] #1 -- #2 ; \end{scope} }  \begin{tikzpicture}[scale=1,font=\small] \tkzInit[xmin=0,xmax=12,ymin=0,ymax=.3,ystep=.05] % Draw coordinates \draw[&gt;=stealth', &lt;-&gt;] (0,6) node[above] {$y$} -- (0,0) --  (12.5,0) node[right] {$x$}; % Draw functions and areas \tkzFct[name plot=A,thick,color=red,domain=0:12]{1/(1.5*sqrt(2*pi))*exp(-((x-4.5)**2)/(2*1.5**1))} \tkzDrawArea[opacity=.3,color=blue,domain = 7:12] \tkzFct[name plot=B,thick,color=blue,domain=0:12]{1/(2*sqrt(2*pi))*exp(-((x-7)**2)/(2*2**1))} \tkzDrawArea[opacity=.3,color=red,domain=0:7] \tkzDrawAreafg[between=b and a,opacity=.3,color=green,domain = 0:7] % Intersection between curves \path [name intersections={of=A and B,by=C}]; % Extract coordinates of C \gettikzxy{(C)}{\cx}{\cy} % Vertical lines \draw [thick,dashed, black] (\cx,0) -- (\cx,5.5) node [above] {$x_{0}$}; \draw [thick,dashed, black] (7,0) -- (7,5.5) node [above] {$\hat{x}$}; % Define regions \Dimline[($(0,0)+(0,-.6)$)][($(7,0)+(0,-.6)$)][above,black][$\mathcal{R}_{1}$]; \Dimline[($(7,0)+(0,-.6)$)][($(12,0)+(0,-.6)$)][above, black][$\mathcal{R}_{2}$]; \end{tikzpicture}  \end{document}
99893
Prove,by vector method,that the point of intersection of the diagonals of the trapezium lies on the line passing through the mid-points of the parallel sides.    My Attempt: Let the trapezium be $OABC$ and that the O is a origin and the position vectors of $A,B,C$ be $\vec{a},\vec{b},\vec{c}$.Then the equation of $OB$ diagonal is $\vec{r}=\vec{0}+\lambda \vec{b}................(1)$ And equation of $AC$ diagonal is $\vec{r}=\vec{a}+\mu(\vec{c}-\vec{a}).......(2)$ And the equation of the line joining the mid points of $OA$ i.e.$\frac{\vec{a}}{2}$ and $BC$ i.e. $\frac{\vec{b}+\vec{c}}{2}$ is $\vec{r}=\frac{\vec{a}}{2}+t(\frac{\vec{b}+\vec{c}}{2}-\frac{\vec{a}}{2}).........(3)$. Here $\lambda,\mu,t$ are scalars. I do not know how to solve $(1)$ and $(2)$ and put into $(3)$ to prove the desired result. Please help me.Thanks.
56090
Can someone please help me solve this problem? I am a bit confused.  Embed $R^2$ in the projective plane $RP^2$ by the map $(x, y) → [1, x, y]$. Find the point of intersection in $RP^2$ of the projective lines corresponding to the parallel lines $y = mx$ and $y = mx + c$ in $R^2$ (where $c \neq  0$).  My attempt: So this transformation sends each point in $R^2$ to the line through the origin which is generated by the vector $(1,x,y)$. Then the line $y=mx$ will go to the set   $[1,x,mx],  x \in \Re, x \neq 0 $.  By the same reasoning, $y=mx+c$ will go to the set {$[1,x,mx+c],  x \in \Re, x \neq 0 $}  The problem is I can't quite understand how to define the points at infinity.I don't want to just do the question, I'd like to understand the basics and will be very grateful if someone could help me.
5971
I have two arbitrary lines in 3D space, and I want to find the distance between them, as well as the two points on these lines that are closest to each other.  Naturally, this only concerns the skew case, since the parallel and intersecting cases are trivial.  I know how to find the distance, as the question was asked before and answered .  I haven't found a good explanation on how to find the two points that determine that distance, though.  So specifically, given two lines  $$L_1=P_1+t_1V_1$$ $$L_2=P_2+t_2V_2$$  I would like to find two points $X_1$ on $L_1$ and $X_2$ on $L_2$ such that the distance between $X_1$ and $X_2$ is minimal.
115265
Problem:    How many triangles do $m$ lines form if   a) Every two lines intersect and no three lines intersect at one point.  b) There are $n$ lines among $m$ lines that are parallel to each other. No other line is parallel to these $n$ lines, and no other two lines are parallel to each other. Again no three lines intersect at one point.  Thank you.
69783
Here is my construction,     If e1 or e2 are positioned a non-zero angle with respect to f, a   single intersection point is directly considered. If e1 is unfixed,   the sweeping of e1 through the intersection point is allowed.   My question is as I am trying to use only one edge (e1 and e2 are edges), in the above are should be is or not
52006
How can I Tex this equation?  I try to have exactly the same result but for the moment I only succeeded in having intersecting lines... Thanks!
79425
Its very common in illustration to want to draw a line towards a vanishing point that is off of the page.   The specific problem is this: lets say we draw two line segments on a piece of paper lying on a flat surface. The line segments are drawn such that, if extended beyond the edge of the page, they intersect at a point $A$. Next, we draw a point $B$ anywhere on the page.   Without actually extending the two line segments to find $A$, is there a way to draw a line segment such that, if extended to a line, would intersect both $A$ and $B$ using only a straight edge and compass construction?   Looking for an answer with the fewest number of steps.
70756
I have a situation like this:     What I need to do is to connect each point to every line that is at most, let's say 200 m, away from the point. In other words, I need to draw a perpendicular line from each point to every line that is in the buffer.  Is there a way to do this in PyQGIS?
209
How can I split all the lines in a feature class where they intersect with a second line feature.  I have a roads feature and a rail feature. I want the lines in the rail feature to split where ever it crosses a road. I'm looking to do it all at once with an overlay analysis tool. Like the Split Line at Point tool, but using lines as the input. Too many to do it manually.
21997
I am having some issues with trying to detect collision of two isometric tile.  I have tried plotting the lines between each point on the tile and then checking for line intercepts however that didn't work (probably due to incorrect formula)  After looking into this for awhile today I believe I am thinking to much into it and there must be a easier way.  I am not looking for code just some advise on the best way to achieve detection of overlap
81253
I know how to find the distance between a point and a line, not between two lines.  Find the shortest distance between the lines $(-1,1,4) + t(1,1,-1)$ and $(5,3,-3) + s(-2,0,1)$  Any help would be appreciated.
58310
I am trying to find the line through the points $(0 : 1 : 0)$ and $(1 : 1 : 1)$ in $\mathbb P^2$ and $(0 :1 : 0: 1)$ and $(1: 1: 1: 0)$ in $\mathbb P^3.$  Would the first line be the set of points $\{(b, a+b, b)\}$ for all $a,b \in \mathbb P^1$  and the second, $\{(d, c+d, d,c)\}$ for all $c, d \in \mathbb P^1$?  Thanks
59379
Am trying to draw a simple rectangle and there seems to be minor issue with two of the endpoints (f and d in the following). I tried line cap=rect as suggested by  but that did not seem to work.  The resulting image:    \documentclass{standalone} \setlength\PreviewBorder{5pt}%  \usepackage{tikz}  \newcommand*{\Width}{2.0} \newcommand*{\Height}{3.0} \newcommand*{\Depth}{1.0}  \begin{document} \begin{tikzpicture}%     \coordinate (A) at ( 0,       0,     \Depth);%     \coordinate (B) at (\Height,  0,     \Depth);%     \coordinate (C) at (\Height,  0,      0);%     \coordinate (D) at (\Height, \Width,  0);%     \coordinate (E) at ( 0,      \Width,  0);%     \coordinate (F) at ( 0,      \Width, \Depth);%     \coordinate (G) at (\Height, \Width, \Depth);%     \coordinate (H) at ( 0,       0,      0);%     %     \draw [blue, ultra thick,line cap=rect]%         (A) -- (B) -- (C) -- (D) -- (E) -- (F) -- cycle;%     \draw [blue, ultra thick,line cap=rect]%         (B) -- (G) -- (F) -- (E) -- (D) -- (G) -- cycle;%      \path (D)  [right] node {$d$};% Problem Point 1     \path (F)  [left ] node {$f$};% Problem Point 2 \end{tikzpicture}% \end{document}   This problem about  seems related, but I do not understand how to adapt that to my problem.
97537
How can I create a straight line when I know the coordinates of the first point, and I know the distance and angle to the end point but I don't know the coordinates of the end point?   Example:    Start point 1: -44.14984703 -19.82517271  End point 2: (unknown lat and long) distance => 200km (note: at 75 degrees from point 1)   Note: if there is also a way in which I can discover the lat and long based on the distance from point 1 and the degree (75º), I think it would work too, I just don't know how to apply.  (Translated from Portuguese using .)