Question

A perpendicular bisector, CD is drawn through point C on AB. 

Answer 1

Answer:

The x-intercept of CD is B(18/5,0). The point C(32,-71) lies on the line CD.

Step-by-step explanation:

Given information: CD is perpendicular bisector of AB.

The coordinates of point A are (-3, 2) and the coordinates of point B are (7, 6).

Midpoint of AB is C.

$C=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})=(\frac{7-3}{2},\frac{2+6}{2})=(2,4)$

The coordinates of C are (2,4).

The slope of line AB is

$m_1=\frac{y_2-y_1}{x_2-x_1}=\frac{6-2}{7-(-3)}=\frac{4}{10}=\frac{2}{5}$

The product of slopes of two perpendicular lines is -1. Since the line CD is perpendicular to AB, therefore the slope of CD is

$m_2=-\frac{5}{2}$

The point slope form of a line is

$y-y_1=m(x-x_1)$

The slope of line CD is $-\frac{5}{2}$ and the line passing through the point (2,4), the equation of line CD is

$y-4=-\frac{5}{2}(x-2)$

$y=-\frac{5}{2}x+5+4$

$y=-\frac{5}{2}x+9$       .... (1)

The equation of CD is $y=-\frac{5}{2}x+9$.

Put y=0, to find the x-intercept.

$0=-\frac{5}{2}x+9$

$\frac{5}{2}x=9$

$x=\frac{18}{5}$

Therefore the x-intercept of CD is B(18/5,0).

Put x=-52 in equation (1).

$y=-\frac{5}{2}(-52)+9=139$

Put x=-20 in equation (1).

$y=-\frac{5}{2}(-20)+9=59$

Put x=32 in equation (1).

$y=-\frac{5}{2}(32)+9=-71$

Put x=-54 in equation (1).

$y=-\frac{5}{2}(-54)+9=144$

Only point (32,-71) satisfies the equation of CD. Therefore the point C(32,-71) lies on the line CD.

Answer 2

Answer

B(18/5,0)

Explanation

First we find the coordinates of C;

C (x, y) = [(-3+7)/2, (2+6)/2]

           = (2, 4)

Find the equation of CD.

slope = (6-2)/(7--3)

         = 4/10  = 2/5

slope of CD = -5/2

-5/2 = (y - 4)/(x - 2)

-5/2(x - 2) = y - 4

(-5/2) x + 5 = y - 4

y = (-5/2)x + 9

For the x-intercept y = 0

∴ y = (-5/2)x + 9  

 0 = (-5/2)x + 9  

5/2 x = 9

x = 2/5 × 9

  = 18/5

x-intercept  = (18/5, 0)