Absolutely clueless on how to do this... Geometry Help

NascentOxygen

The general equation for a straight line is y=mx+c
You seem to be overlooking that x.

As I pointed out, you now have all the information you need to write the equation for every perpendicular bisector.

For each you know its slope, and you know the co-ordinates of a point on the side of the triangle that it passes through. So, considering one perpendicular at a time, substitute for what you know into the straight line equation y=mx+c and hence determine its equation.

SammyS

Find the equation for each of the lines corresponding to these bisectors.

For the perpendicular bisector to the segment with vertices (-a,0) and (b,c):

[itex]\displaystyle y=-\frac{b+a}{c}\,x+k_1[/itex]

Put in the coordinates for the midpoint to find k₁

k₁ is the y-intercept for this line, so that's where this line intersects the line, x=0.

Do similar for the other bisector.

Frogeyedpeas

Okay so lets use the first perpendicular bisector: and the corresponding midpoint

((b-a)/2, c/2)

We know that y = -(b+a)/c*x + K1 So to substitute the x value we get

c/2 = -(b²-a²)/c + K1

multplying the c we recieve:

c²/2 = k -b² + a²

Which leaves us finally with

K = 1/2(c²+b²-a² = k

And since that is the Y coordinate of the point of intersection we get the intersection of three perpendicular bisectors as:

(0, 1/2c² + b²-a²) which unfortunately doesn't match up with my book's answer (I wanted to do this from scratch without working backwards) which is:

(0, (-a² + b²+²)/2c)

The two expressions are not equal either, yet this one that I have attained seems quite logically deduced

SammyS

Not quite right.

You failed to multiply k by c. Anyway there should have been a 2 in both denominator.

So, after multiplying by 2c you should have:

[itex]\displaystyle c^2=2ck-b^2+a^2[/itex]

Solving for k gives:

[itex]\displaystyle k=\frac{b^2+c^2-a^2}{2c}[/itex]