Question

A(7-1), B(6, 4) and C(5, 5) are three

Answer 1

Answer:

The answer is below

Step-by-step explanation:

The equation of the line passing through two points is given by:

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

The equation of line AB is:

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\\\\y-(-1)=\frac{4-(-1)}{6-7}(x-7)\\\\y+1=-5(x-7)\\y+1=-5x+35\\y=-5x+34$

The midpoint of two lines is given as:

$x=\frac{x_1+x_2}{2}, y=\frac{y_1+y_2}{2}\\midpoint\ of\ AB \ is:\\x=\frac{7+6}{2}=6.5, y=\frac{-1+4}{2}=1.5\\ = (6.5,1.5)$

The equation of line AC is:

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\\\\y-(-1)=\frac{5-(-1)}{5-7}(x-7)\\\\y+1=-3(x-7)\\y+1=-3x+21\\y=-3x+20$

The midpoint of two lines is given as:

$x=\frac{x_1+x_2}{2}, y=\frac{y_1+y_2}{2}\\midpoint\ of\ AB \ is:\\x=\frac{7+5}{2}=6, y=\frac{-1+5}{2}=2\\ = (6,2)$

The product of the slope of a perpendicular bisector of a line and the slope of the line is -1. That is m1m2 = -1

The slope of the perpendicular bisector of AB is:

m(-5)=-1

m=1/5

The equation of the perpendicular bisector of AB passing through (6.5,1.5) is:

$y-y_1=m(x-x_1)\\y-1.5=\frac{1}{5}(x-6.5)\\y=\frac{1}{5} x-8.25$

The slope of the perpendicular bisector of AB is:

m(-3)=-1

m=1/3

The equation of the perpendicular bisector of AB passing through (6,2) is:

$y-y_1=m(x-x_1)\\y-2=\frac{1}{3}(x-7)\\y=\frac{1}{3} x-4.33$

2) The point of intersection is gotten by solving y = 1/5 x -8.25 and y = 1/3 x-4.33 simultaneously.

Subtracting the two equations from each other gives:

0= -0.133x - 3.92

-0.133x = 3.92

x = -29.5

Put x = -29.5 in y = 1/5 x -8.25 i.e:

y = 1/5 (29.5) -8.25

y = -14.16

The point of intersection is (-29.5, -14.16)