Given three points A(-1,-2) B(3,4) C(3,6). Find the perpendicular bisector of a) AB b)BC​

The perpendicular bisector is line that divides another line into two equal parts, and is also at right angles to the line it is bisecting.Question A) We need to work out the following: Midpoints of AB (since it's a bisector) Perpendicular gradient of AB (since its the perpendicular bisector) Finally, equation of perpendicular gradient, through the midpoint of AB (since the bisector is halfway through AB)Working out the Midpoint of AB:To work out the midpoint, you take the average (add and divide by 2) of the x coordinates and the y coordinates for points A and B:xcoords of Midpoint = [tex]\frac{-1+3}{2}}=1[/tex]ycoords of Midpoint = [tex]\frac{-2+4}{2} =1[/tex]So the coordinates of the Midpoint is: (1, 1)Working out perpendicular gradient:First, we work out the gradient of AB. Then to get the perpendicular gradient, we take the negative reciprocal of that gradient (by flipping the fraction and putting a negative sign in front).[tex]Gradient =\frac{y1-y2}{x1-x2}=\frac{-2-4}{-1-3}=\frac{-6}{-4}=\frac{3}{2}[/tex]So perpendicular gradient = negative reciprocal of [tex]\frac{3}{2}[/tex]=[tex]-\frac{2}{3}[/tex]Equation of Perpendicular gradient through MidpointTo get the equation, we can use the following formula:[tex]y-y1=m(x-x1)[/tex]Where:y1 = y coordinate of the midpoint we worked out  (1 )x1 = x coordinate of the midpoint we worked out  (1 )m = perpendicular gradient we worked out           (-2/3)So all we do now is substitute in the values into the equation and rearrange for y:[tex]y -y1=m(x-x1)[/tex][tex]y-1=-\frac{2}{3} (x-1)[/tex]   (times both sides by 3)[tex]3y-3=-2(x-1)[/tex]   (expand brackets)[tex]3y -3=-2x+2[/tex]   [tex]3y=-2x +5[/tex][tex]y=-\frac{2}{3}x+\frac{5}{3}[/tex]Answer:Perpendicular Bisector of AB is: [tex]y=-\frac{2}{3}x+\frac{5}{3}[/tex] ____________________________________Question B)We just follow the same steps as for Question A, except we use the coordinates for B and C, instead of A and B, so I will include less explanation for this question.[tex]Midpoint=(\frac{x1+x2}{2},\frac{y1+y2}{2})=(\frac{3+3}{2},\frac{4+6}{2})=(3, 5)[/tex]Gradient:Notice, the x coordinates for B(3, 4) and C (3, 6) are the same. That means the gradient is 0, and so is the perpendicular gradient. So we can skip straight to get the equation of perpendicular bisector.Equation of perpendicular bisector:[tex]y-y1=m(x-x1)[/tex][tex]y-5=0(x-3)[/tex][tex]y-5 = 0[/tex][tex]y=5[/tex] Answer:The perpendicular bisector of BC is:  [tex]y=5[/tex] _______________________________________________Answers:a) [tex]y=-\frac{2}{3}x+\frac{5}{3}[/tex] b) [tex]y=5[/tex]