17. the distance between (2, 1) and (n, 4) is 5 units. find all possible values of n.

6=n   and -2=n

Step-by-step explanation:

The distance between two points is found by

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

5 = sqrt(( n-2)^2 + (4-1)^2)

5 = sqrt(( n-2)^2 + (3)^2)

Square each side

5^2 = sqrt(( n-2)^2 + (3)^2)^2

25 = ( n-2)^2 + (3)^2

25 = (n-2)^2 +9

Subtract 9 from each side

25-9 = (n-2)^2 +9-9

16 = (n-2)^2

Take the square root of each side

sqrt(16) = sqrt( (n-2)^2)

±4 = n-2

Add 2 to each side

2±4 = n-2+2

2 ±4 = n

2+4 =n   and   2-4 =n

6=n   and -2=n

where's the diagrams?

the sum of the lengths of any two sides of a triangle is greater than the length of the third side. if you take the three sides of a triangle and add them in pairs, the sum is greater than (not equal to) the third side. if that is not true, then it is not possible to construct a triangle with the given side lengths.

step-by-step explanation: