distance between points

Question

If AB=CD then find t

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

Answer 1

Lets

A(x₁,y₁)=(t,1)

B(x₂,y₂)=(2,3)

Distance between two points A(x₁,y₁) and B(x₂,y₂) is given by

d=√(x₂-x₁)²+(y₂-y₁)²

  =√(2-t)²+(3-1)²

  =√(2)²-2(2)(t)+t²+(2)²                     {Recall:(a-b)²=a²-2ab+b²}

  =√4-4t+t²+4

d=√t²-4t+8

Lets

C(x₁,y₁)=(7,0)

D(x₂,y₂)=(0,0)

Distance between two points A(x₁,y₁) and B(x₂,y₂) is given by

d=√(x₂-x₁)²+(y₂-y₁)²

  =√(0-7)²+(0-0)²

  =√(-7)²+0                                    

  =√49

d=7

The condition is AB=CD

√t²-4t+8 = 7

Takes squares each side

(√t²-4t+8)² = 7²

Cancel the squares and roots

t²-4t+8 = 49

Subtract each side by 49

t²-4t+8-49 = 49-49

t²-4t-41 = 0                                                (Subtract:49-49=0, 8-49=-41)

The quadratic formula is

x= (4±(√16+164))/2                              ((-4)2=16, (-4)(-41) = 164 )

x = (4±(√180))/2 

x = (4±2√45)/2                                        (√(4×45)=2 √45)

x = 2(2±√45)/2                                        (Divide:2/2=1)

x=(2±√45)

x=(2+√45)  or  x =    (2-√45)

The value of x  is (2+√45) ,(2-√45).