1. ## perpendicular bisector

the equations of perpendicular bisectors of the sides AB and AC of a triangle ABC are x-y+5=0 and x+2y=0 respectively.If this point A(1,-2) the equation of the line BC is

2. ## Re: perpendicular bisector

Originally Posted by prasum
the equations of perpendicular bisectors of the sides AB and AC of a triangle ABC are x-y+5=0 and x+2y=0 respectively.If this point A(1,-2) the equation of the line BC is
Step 1. Find the point, D, where those two lines intersect.
Step 2. Find the distance, d(A,D). The is the circum-radius.
Step 3. Write the equations of the two lines through A each of which is perpendicular to one of the given lines
Step 4. Find the points on the lines in step 3 where the circum-circle insects.
Step 5: Those points are B & C.

3. ## Re: perpendicular bisector

Originally Posted by prasum
the equations of perpendicular bisectors of the sides AB and AC of a triangle ABC are h: x-y+5=0 and g: x+2y=0 respectively.If this point A(1,-2) the equation of the line BC is
1. Since $AC \perp g$ the line AC has the slope $m_{AC} = 2$. Use point-slope-formula to determine the equation of

$AC: y+2=2(x-1)$

2. $AC\ \cap g = M_{AC}$

$2x-4 = -\frac12 x~\implies~x=\frac85$

Consequently the midpoint of $\overline{AC}$ is $M_{AC}\left(\frac85\ ,\ -\frac45\right)$

Now you can determine the coordinates of point C: $C\left(\frac{11}5\ ,\ \frac25\right)$

3. The same method will give you the point B: $B(-7, 6)$

4. Use the coordinates of B and C to determine the equation of BC.

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# perpendicular bisector of AB and AC are x-y 5=0 and x 2y=0 pount A(1,-2)

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