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Thread: finding cordinate of a point ( i know distance and gradient)

  1. #1
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    finding cordinate of a point ( i know distance and gradient)

    i have the the cordinate of a point

    i have the distance of a line

    i have the gradient of the line

    i want to know the cordinate of the 2nd point

    that is from the first cordinate, add the distance to it considering the gradient

    i need to know the 2nd cordinate

    i know i might get 2 cordinates but it dont matter

    how to do it can someone help?
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  2. #2
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    The gradient (slope) of a line is the tangent of the angle, $\displaystyle \theta$ the line makes with the x-axis. And $\displaystyle tan(\theta)= \frac{length of opposite side}{lenght of near side}= \frac{\Delta y}{\Delta x}$. If the gradient is m then $\displaystyle m= \frac{\Delta y}{\Delta x}$ or, equivalently, $\displaystyle m\Delta x= \Delta y$.

    And, of course, if length of the line is d, then $\displaystyle d^2= \Delta x^2+ \Delta y^2$ so that $\displaystyle \Delta y^2= d^2- \Delta x^2$ and $\displaystyle \Delta y= \sqrt{d^2- \Delta x^2}$.

    Putting that into the equation for the gradient, m, $\displaystyle m\Delta x= \sqrt{d^2- \Delta x^2}$.

    Square both sides to get $\displaystyle m^2\Delta x^2= d^2- \Delta x^2$ which is the same as $\displaystyle (m^2+ 1)\Delta x^2= d^2$.

    Solve that for $\displaystyle \Delta x$ and then find $\displaystyle \Delta y$ from $\displaystyle \Delta y= m\Delta x$

    $\displaystyle \Delta x$ and $\displaystyle \Delta y$ are the changes in the x and y coordinates to add then to the x and y values of the initial point to find the final point.

    Yes, there are two possible answers- the points on the same line on either side of the initial point. When you solve $\displaystyle (m^2+ 1)\Delta x= d^2$, you will need to take a square root and which point you get depends on which sign you take for $\displaystyle \Delta x$.
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