+0

# In the Diagram A is the point (-1,1) B is (3,3) and C is (6,2).

0
49
3

In the Diagram A is the point (-1,1) B is (3,3) and C is (6,2).

The perpendicular bisector of AB has equation y + 2x = 4

(A)Find the equation of the perpendicular bi-sector of BC.

(B) Find the centre of the circle wich passes through A,B & C

For A I have worked out that the gradient for the bisector of BC is 3

I have also worked out the mid point (9/2 , 5/2).

However when I put this into the form y-b=m(x-a) it doesn't work out properly.

Guest Sep 4, 2017
Sort:

#1
+1114
+1

For A, I think I know why it is not working properly. You appear to have done everything correctly, so I presume that there is a silly computational error.

1Determine the Midpoint of Both Coordinates

The midpoint is the average of both x- and y- coordinates. In algabraic terms, the formula is $$\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$$. Let's use it! You used this formula correctly, as I can tell. For completeness, I'll calculate it also.

 $$\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$$ Now, just plug into the formula. $$\left(\frac{3+6}{2},\frac{3+2}{2}\right)$$ Now, simplify both sides. $$\left(\frac{9}{2},\frac{5}{2}\right)$$

2. Determine Slope (or Gradient) of the Segment

This also has a formula, luckily, and all we have to do is plug the coordinates in. It is $$m=\frac{y_2-y_1}{x_2-x_1}$$, where m is the slope of the segment. Let's just use the formual to determine the slope.

 $$m=\frac{y_2-y_1}{x_2-x_1}$$ When using this formula, it is essential to plug in the x- and y-coordinates in the same order. Otherwise, you will get the incorrect result. $$m=\frac{3-2}{3-6}$$ Now, simplify the numerator and denominator. $$m=-\frac{1}{3}$$

This is not the slope of the perpendicular bisector but rather the slope of the original 2 segments.

3. Determine The Slope of Perpendicular Bisector

A line that is perpendicular to a segment has a slope that is opposite and reciprocal to the slope of the original segment. What does that mean? Well, let's deal with the current slope as its example of $$-\frac{1}{3}$$.

I find doing the reciprocal of a number first is generally easier, but the order in which you do the next steps do not matter. In general, the reciprocal of a number follows tthe following rule of $$a\Rightarrow\frac{1}{a}$$. Let's apply that now with our slope. $$-\frac{1}{3}\Rightarrow-\frac{1}{\frac{1}{3}}=-3$$

Now, find the opposite. To do this, use the rule that $$a\Rightarrow-a$$$$-3\Rightarrow3$$.

Therefore, $$3$$ is the slope of the perpendicular bisector.

If you are interested, here is another method to obtain the slope of the perpendicular bisector. If you take the original slope of the segment and multiply it by the slope of the perpendicular bisector, you will get -1. Let's see this in action!

 $$-\frac{1}{3}m=-1$$ Multiply by -3 on both sides. $$m=3$$

Look at that! you get the same answer.

4. Use Point-Slope Form

There are 2 methods again to get to the final equation of the slope, but I generally like point-slope form for some reason, but I will demonstrate both to you, so that you can choose your method. Point-slope form is the following:

$$y-y_1=m(x-x_1)$$ where $$(x_1,y_1)$$ is a point on said segment or line and m is the slope of said segment or line. We know what segment is on this perpendicular bisector; we already calculated it. The perpendicular bisector is a segment that both intersects a segment at its midpoint and at a perpendicular angle. We have already calculated the midpoint of BC, which is $$\left(\frac{9}{2},\frac{5}{2}\right)$$. We also know the slope, which is 3. Since we have all that information, let's plug it in!

 $$y-y_1=m(x-x_1)$$ As aforementioned, plug all the information we know into the equation. $$y-\frac{5}{2}=3\left(x-\frac{9}{2}\right)$$ Now, we are transforming this equation to slope-intercept form. To do that, let's distribute the 3 to both sides. $$y-\frac{5}{2}=3x-\frac{27}{2}$$ We know that we must do some manipulation with the fraction. $$y-\frac{5}{2}=3x+\frac{-27}{2}$$ Now, add $$\frac{5}{2}$$ $$y=3x-11$$

This is the equation of the perpendicular bisector. I'm not sure where you went wrong, unfortunately, but it seems to me that it is probably some silly computational error. That's all.

Another method of doing this is using the slope-intercept form from the beginning. Let's see how that would work:

 $$y=mx+b$$ We know what the slope is; it is 3, so let's plug that in. $$y=3x+b$$ Now, plug in a point that we know is on the perpendicular bisector. We know that $$\left(\frac{9}{2},\frac{5}{2}\right)$$ is a point, so plug it in. $$\frac{5}{2}=\frac{3}{1}*\frac{9}{2}+b$$ Now, solve for b. $$\frac{5}{2}=\frac{27}{2}+b$$ Subtract 27/2 on both sides. $$b=-\frac{22}{2}=-11$$ Therefore, the equation becomes the following. $$y=3x-11$$ This should look very familiar to what we determined above.

Hopefully, this helped to find the segment that is the perpendicular bisector.

Now, it is time to do B. Thankfully, this will not be as time-consuming as generating this repsonse has taken me hours. The point you want me to identify is called the circumcenter of the triangle. it is the point located at the center of a circle such that circle contains all the points in a triangle. The circumcenter of a triangle can be found by finding the intersection of 2 of the perpendicular bisectors! Let's do that. We already know 2 lines and their respective equations. They are $$y=3x-11$$ and $$y+2x=4$$. Let's set up a system of equations.

$$y=3x-11$$

$$y+2x=4$$

I will use substitution in this case as we already know what y equals.

 $$3x-11+2x=4$$ Combine the like terms on the left-hand side. $$5x-11=4$$ Add 11 to both sides. $$5x=15$$ Divide by 5 on both sides to get the corresponding x-coordinate. $$x=3$$

Now, plug in x into one of the equations. I think that equation 1 will be easier, so I will use that!

 $$y=3x-11$$ Plug in for x, which is 3. $$y=3*3-11=9-11=-2$$

Therefore, the circumcenter of the triangle is $$(3,-2)$$

If you would like, click here to view a graph of the triangle, its respective lines that are perpendicular bisectors and the circumcenter.

TheXSquaredFactor  Sep 4, 2017
edited by TheXSquaredFactor  Sep 4, 2017
edited by TheXSquaredFactor  Sep 4, 2017
#2
+4484
+2

I didn't know that about the circumcenter

Also...thanks for the detailed graph...it's very helpful !

hectictar  Sep 4, 2017
#3
+1114
+2

I have updated the graph such that the three vertices of the triangle lie on the circle.

I have done this to demonstrate what the circumcenter is a tad clearer (if it wasn't already!). As I said before, the circumcenter is the center of a circle where those 3 vertices of a triangle lie on the circle.

The more you know... On the other hand, I'm close to reaching that 1000 ♥ benchmark. Can't wait!

TheXSquaredFactor  Sep 4, 2017
edited by TheXSquaredFactor  Sep 4, 2017

### 20 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details