Question

The distance between the points(-3, -5) and (2,7) on the coordinate plane is given as 5^2+ t^2. what us the value of t​

Answer 1

Explanation:

so we can use
`t^(2)` + 25 as one expression and the find the distance between the points and set the expression equal to that number. That's our strategy.

dist = the square root of
`x^(2)` +
`y^(2)`

where x and y are the change in x2 to x1 and y2 to y1

P1 = ( -3, -5)

P2 = (2, 7 )

dist = square root of ( [2 -(-3) ]^2 + [7-(-5)]2)

dist = sq rt ( 5^2 + 12^2 )

dist = sq rt ( 25 + 144)

dist = sq rt (169)

dist = 13

now set our eq. equal to that number

`t^(2)` + 25 = 13

`t^(2)` = -12

this is looking kinda bad, do you know complex numbers?

t = 0 + j 3.46410

hmmm I'm wondering if you've left out .. or forgotten some thing in the question? :/