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Math Help - Equation of line

  1. #1
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    Equation of line

    Find the equation of the line through M (-1,2) with gradient m if M is the midpoint of the intercepts of the x axis and y axis

    equation of line = y - y1 = m ( x - x1)

    what next I am confused with the midpoint of axis.

    Thanks
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  2. #2
    Super Member Bacterius's Avatar
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    Hello Joel,
    your line passes through the point M (-1, 2), so you know that 2 = -m + p (using the standard equation y = mx + p)
    Now, let x_1 be the intercept of the y-axis, and y_1 be the intercept of the x-axis. Then we have M \left (\frac{x_1}{2}, \frac{y_1}{2} \right ). Thus we have x_1 = -2 and y_1 = 4. Since these are intercepts, we can get some more information :

    0 = -2m + p (using y = mx + p)

    Now recover the equation we found first :

    2 = -m + p

    Solving the two equations simultaneously :

    \left \{<br />
\begin{array}{l}<br />
0 = -2m + p \\<br />
2 = -m + p<br />
\end{array}<br />
\right.

    \left \{<br />
\begin{array}{l}<br />
2m = p \\<br />
2 = -m + p<br />
\end{array}<br />
\right.

    \left \{<br />
\begin{array}{l}<br />
2m = p \\<br />
2 = -m + 2m<br />
\end{array}<br />
\right.

    \left \{<br />
\begin{array}{l}<br />
2m = p \\<br />
2 = m<br />
\end{array}<br />
\right.

    \left \{<br />
\begin{array}{l}<br />
4 = p \\<br />
2 = m<br />
\end{array}<br />
\right.

    \left \{<br />
\begin{array}{l}<br />
p = 4 \\<br />
m = 2<br />
\end{array}<br />
\right.

    And the equation of the line follows : \boxed{y = 2x + 4}

    Plotting this line indeed shows that it passes through M(-1, 2), and that this point is exactly midpoint between the x-intercept and y-intercept of the line

    Don't hesitate to reply if anything seems confusing.
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  3. #3
    Super Member
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    Hi bacterius,
    After reading your first paragraph I accepted this as your proof of the x and y intercepts of the required line ie that the x intercept is -2 and y intercept is 4. So immediately y=2x+4. Was it necessary to go further. I did it by taking a straight edge and rotating it about M so that M was the midpoint between the x and y axis


    bjh
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  4. #4
    Super Member Bacterius's Avatar
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    Indeed, this is a shortcut, since we know the y-intercept we immediately know p, then we can use the point M (not the x-intercept, sorry typo) to solve for m, the gradient. I just did it the long way, I'm glad you noticed it was possible to go faster
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