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
Hello Joel,
your line passes through the point $\displaystyle M (-1, 2)$, so you know that $\displaystyle 2 = -m + p$ (using the standard equation $\displaystyle y = mx + p$)
Now, let $\displaystyle x_1$ be the intercept of the y-axis, and $\displaystyle y_1$ be the intercept of the x-axis. Then we have$\displaystyle M \left (\frac{x_1}{2}, \frac{y_1}{2} \right )$. Thus we have $\displaystyle x_1 = -2$ and $\displaystyle y_1 = 4$. Since these are intercepts, we can get some more information :
$\displaystyle 0 = -2m + p$ (using $\displaystyle y = mx + p$)
Now recover the equation we found first :
$\displaystyle 2 = -m + p$
Solving the two equations simultaneously :
$\displaystyle \left \{
\begin{array}{l}
0 = -2m + p \\
2 = -m + p
\end{array}
\right.$
$\displaystyle \left \{
\begin{array}{l}
2m = p \\
2 = -m + p
\end{array}
\right.$
$\displaystyle \left \{
\begin{array}{l}
2m = p \\
2 = -m + 2m
\end{array}
\right.$
$\displaystyle \left \{
\begin{array}{l}
2m = p \\
2 = m
\end{array}
\right.$
$\displaystyle \left \{
\begin{array}{l}
4 = p \\
2 = m
\end{array}
\right.$
$\displaystyle \left \{
\begin{array}{l}
p = 4 \\
m = 2
\end{array}
\right.$
And the equation of the line follows : $\displaystyle \boxed{y = 2x + 4}$
Plotting this line indeed shows that it passes through $\displaystyle M(-1, 2)$, 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.
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
Indeed, this is a shortcut, since we know the y-intercept we immediately know $\displaystyle p$, then we can use the point M (not the x-intercept, sorry typo) to solve for $\displaystyle m$, the gradient. I just did it the long way, I'm glad you noticed it was possible to go faster