Answer:
The following steps are needed:
1. Find the function of AB with A(-3,2) and B(7,6)
y(AB) = mx + b. Calculate m = (y₂-y₁)/(x₂-x₁) = (6-2)/(7+3) = 4/10 → 2/5
y(AB) = (2/5).x + b. Calculate b; 6 = (2/5).(7) + b and b = 16/5
y(AB) = (2/5).x +16/5
2. calculate the function y(CD) = mx + bWe know already m = -5/2, since CD is perpendicular to AB and hence the product of their slopes = -1
y(CD) = -(5/2).x + b
We also know that y(CD) passes in the middle of AB, then let's calculate the coordinate of C, the midpoint of AB:x=(x₁+x₂)/2 and y=(y₁+y₂)/2)x= (-3+7)/2 and y=(2+6)/2x= 2 and y=4y(CD) = (-5/2).x +b4 = (-5/2).(2)+b and b = 9y(CD) =(-5/2)x + 9
x intercept when y = 0 → 0= (-5/2)x +9 → x= 18/5So x intercept (18/5,0)
So the answer is at point B.
Answer:
The x- intercept of CD is B(18/5,0) and point C(32,-71) lies on CD
Step-by-step explanation:
We are given that a perpendicular bisector CD is drawn through point C on AB.
The coordinates of point A are (-3,2)
The coordinates pf point B are (7,6).
We have to find the x-intercept of CD
Slope formula:[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]
By using the formula
Slope of AB=[tex]\frac{6-2}{7+3}=\frac{4}{10}=\frac{2}{5}[/tex]
Slope of CD=[tex]-\frac{1}{\frac{2}{5}}=-\frac{5}{2}[/tex]
When two lines are perpendicular then slope of one line is equal to negative reciprocal of slope of other line.
Mid point of AB
[tex]x=\frac{-3+7}{2}=2,y=\frac{2+6}{2}=4[/tex]
The mid-point of AB is at (2,4)
By using mid-point formula
[tex]x=\frac{x_1+x_2}{2},y=\frac{y_1+y_2}{2}[/tex]
The equation of line CD is passing through the point (2,4) with slope -5/2 is given by
[tex]y-4=-\frac{5}{2}(x-2)[/tex]
By using point slope form:[tex]y-y_1=m(x-x_1)[/tex]
[tex]y=-\frac{5}{2}(x-2)+4[/tex]
Substitute y=0
[tex]-4=-\frac{5}{2}(x-2)[/tex]
[tex]x-2=\frac{4\times 2}{5}[/tex]
[tex]x-2=\frac{8}{5}[/tex]
[tex]x=\frac{8}{5}+2=\frac{8+10}{5}[/tex]
[tex]x=\frac{18}{5}[/tex]
Hence, the x- intercept of CD is (18/5,0).
[tex]y=-\frac{5}{2}(x-2)+4[/tex]
Substitute x=-52
[tex]y=-\frac{5}{2}(-52-2)+4=139[/tex]
Substitute x=-20
[tex]y=-\frac{5}{2}(-20-2)+4=59[/tex]
Substitute x=32
[tex]y=-\frac{5}{2}(32-2)+4=-71[/tex]
Substitute x=-54
[tex]y=-\frac{5}{2}(-54-2)+4=144[/tex]
Hence, point C(32,-71) lies on CD.
