What is the endpoint $P(x,y)$ of a line segment , if I know its starting point: $C(x(1), y(1))$, the gradient $G$, and its length $L$?
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@AlexM. This question is specifically posed in $\mathbb{R}^2$. (By the way, the "gradient" here is what others may call "slope"). – Jun 07 '15 at 17:12
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HINT:
$$(x-x_1)^2+(y-y_1)^2=L^2$$
and $$G=\dfrac{y-y_1}{x-x_1}$$
lab bhattacharjee
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$$\dfrac{x-x_1}1=\dfrac{y-y_1}G$$ $$=\pm\sqrt{\dfrac{(x-x_1)^2+(y-y_1)^2}{1^2+G^2}}=\pm\dfrac L{\sqrt{1+G^2}}$$ – lab bhattacharjee Jun 07 '15 at 16:16
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$$ D = C + L (G/\lVert G\rVert) $$ where $$ \lVert G\rVert= \sqrt{\sum_{i=1}^2 G_i^2} $$ is the length of $G$.
mvw
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Maybe we mean different things? A gradient of a curve $u$ is the vector $(\partial_x u, \partial_y u)$. – mvw Jun 07 '15 at 16:24
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$C$ is the start point you gave in your question, $D$ is the end point of the segment $CD$. – mvw Jun 07 '15 at 16:33
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