Question

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

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:

the x-intercept of CD is[ A(3,0) B(18/5,0) C(9,0) D(45/2,0) ] . Point [ A(-52,117) B(-20,57) C(32,-71) D(-54,-128) ] lies on CD.

Given :

CD is perpendicular bisector of AB.

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

C is the midpoint of AB.

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

The coordinates of C are (2,4).

Line AB has a slope of:$m_1=\frac{y_2-y_1}{x_2-x_1}=\frac{6-2}{7-(-3)}=\frac{4}{10}=\frac{2}{5}$

The product of slopes of two perpendicular lines is -1. Since the line CD is perpendicular to AB, therefore the slope of CD : $m_2=-\frac{5}{2}$

The point slope form of a line is given by:

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

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

$y-4=-\frac{5}{2}(x-2)\\y=-\frac{5}{2}x+5+4\\y=-\frac{5}{2}x+9 .... (1)$

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

In order to find the x-intercept, put y=0.

$0=-\frac{5}{2}x+9\\\frac{5}{2}x=9\\x=\frac{18}{5}$

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

Put x=-52 in eq(1).

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

Put x=-20 in eq(1).

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

Put x=32 in eq(1)

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

Put x=-54 in eq1).

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

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