# Thread: Equation of Line Through C & Midpoint

1. ## Equation of Line Through C & Midpoint

Let A(1,2), B(6,1) and C(7,8) be three points in the plane. Find the equation of the line passing through point C and through the midpoint of the line segment AB. Write answer in the form
ax + by + c = 0.

I found slope of AB to be (-1/5).
I also found the midpoint of AB to be (7/2, 3/2).

I then decided, for unknown reasons, to calculate the slope of the perpendicular bisector to be 5.

I then plugged 5 and the midpoint into point-slope formula and got the wrong answer. Trying a second time through guessing, I decided to plug 5 and point C into the point-slope formula and again got the wrong answer. The textbook does not give a sample representing the question above or else I can do it in 5 minutes.

What are the steps leading to the right equation? I just want the steps to see if I can work it out alone. Thank you very much.

2. ## Re: Equation of Line Through C & Midpoint

Originally Posted by nycmath
Let A(1,2), B(6,1) and C(7,8) be three points in the plane. Find the equation of the line passing through point C and through the midpoint of the line segment AB. Write answer in the form
ax + by + c = 0.
I also found the midpoint of AB to be (7/2, 3/2).
The mid point is $M: \left( {\frac{7}{2},\frac{3}{2}} \right)$.

Now find the slope determined by $C~\&~M$. Then write the line.

3. ## Re: Equation of Line Through C & Midpoint

Notice that C to Midpoint of AB is not necessarily the same as a perpendicular of AB, therefore your entire problem remains.

Let the Midpoint of AB be M(X,Y) and you have already found that answer to be $(\frac{7}{2}, \frac{3}{2})$

Finding the Gradient is easy, since we know two points on the line. $\frac{8-\frac{3}{2}}{7-\frac{7}{2}}=\frac{13}{7}$

Equation of the Line $(y-8)=\frac{13}{7}(x-7)$

$13x-7y-35=0$

Hope this helps

4. ## Re: Equation of Line Through C & Midpoint

Originally Posted by Plato
The mid point is $M: \left( {\frac{7}{2},\frac{3}{2}} \right)$.

Now find the slope determined by $C~\&~M$. Then write the line.
Thank you so much.

5. ## Re: Equation of Line Through C & Midpoint

Originally Posted by LimpSpider
Notice that C to Midpoint of AB is not necessarily the same as a perpendicular of AB, therefore your entire problem remains.

Let the Midpoint of AB be M(X,Y) and you have already found that answer to be $(\frac{7}{2}, \frac{3}{2})$

Finding the Gradient is easy, since we know two points on the line. $\frac{8-\frac{3}{2}}{7-\frac{7}{2}}=\frac{13}{7}$

Equation of the Line $(y-8)=\frac{13}{7}(x-7)$

$13x-7y-35=0$

Hope this helps
I truly appreciate it.