If AB=CD then find t using distance formula

Question

A(t, 1), B(2, 3) C(7, 0) and D(0, 0). thanks in advence..

Answer 1

Given that  A(t, 1), B(2, 3) C(7, 0) and D(0, 0)

First find out AB

A = (t, 1) = (x1,y1), B = (2, 3) = (x2,y2)

AB = sqrt(x2-x1)^2+(y2-y1)^2

AB = sqrt(2-t)^2+(3-1)^2

AB = sqrt(2-t)^2+(2)^2

AB = sqrt(4+t^2-4t)+4

AB = sqrt(8+t^2-4t)

Squaring on each side

AB^2 = (8+t^2-4t)

C = (7, 0) = (x1,y1), D = (0, 0) = (x2,y2)

CD = sqrt(x2-x1)^2+(y2-y1)^2

CD = sqrt(0-7)^2+(0-0)^2

CD = sqrt(-7)^2+(0)^2

CD = sqrt49

Squaring on each side

CD^2 = 49

Given that  AB = CD

Squaring on each side

AB^2 = CD^2

8+t^2-4t = 49

Subtract 49 from each side

8+t^2-4t-49 = 49-49

t^2-4t-41 = 0

Compare the above equation with ax^2+bx+c = 0.so a=1,b= -4and c= -41

t = -b±sqrtb^2-4ac/2a

t = -(-4)±sqrt(-4)^2-4(-41)/2

t = 4±sqrt16+164/2

t = 4±sqrt180/2

t = 4±6sqrt5/2

t = 2±3sqrt5

The values of t = 2±3sqrt5