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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

Guest Mar 18, 2018
 #1
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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 :)

Melody  Mar 18, 2018

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