find the distance between the points (1, 2t) and (1-t,1) is 11-9t, find the possible values of t.

Question

...pls answer

Answer 1

Use the distance formula:

 

d = √[(x₁ - x₂)² + (y₁ - y₂)²]

 

 

d = 11 - 9t

Point 1 = (1 , 2t)

Point 2 = (1 - t , 1)

 

Plug in these values into the formula.

 

11 - 9t = √[(1 - (1 - t))² + (2t - 1)²]

 

11 - 9t = √[(t² + (4t² - 4t + 1)]

 

 

Square both sides of the equation and expand.

 

121 - 198t + 81t² = 5t²- 4t + 1

 

 

Move all the terms to the right side.

 

0 = -76t² + 194t - 120

 

 

Use the quadratic formula to solve for t.

 

t = (-b ± √(b² - 4ac)) / 2a

 

where:

a = -76

b = 194

c = -120

 

Plug in these values, and you are all set.