I’ve tried so hard on this, can anyone do it?
The points A (6,1) and B(-2,5) are on the line with equation y=-1/2x+4
M is the midpoint of AB
Find an equation of the line through M that is perpendicular to y=-1/2x+4
Hi, I can help you :)
The points A (6,1) and B(-2,5) are on the line with equation y=-1/2x+4
M is the midpoint of AB
Find an equation of the line through M that is perpendicular to y=-1/2x+4
the gradient of the line y=-1/2x+4 is \( -\frac{1}{2}\)
The gradient of the line that is perpendicular is the negative reciprocal which is \( +\frac{2}{1}=2\)
So 2 will be the gradient of the new line.
The midpoint of AB is simply the average of the xes and the average of the ys
\(midpoint=(\frac{6+-2}{2},\frac{1+5}{2})\\ midpoint=(\frac{4}{2},\frac{6}{2})\\ midpoint=(2,3)\\\)
So you want the equation of the line through (2,3) with a gradient of 2
there are lots of methods to do this.
Here is one, my start is a bit different from normal, I think it makes it easier because it is the same for many different questions. You just need 2 different ways to express the gradient to make it work.
\(\begin{align} gradient &=gradient\\ m&=\frac{y-y_1}{x-x_1}\\ 2&=\frac{y-3}{x-2}\\ 2(x-2)&=y-3\\ 2x-4&=y-3\\ 2x-1&=y\\ y&=2x-1 \end{align} \)
I hope all that helps :)