Question

The axes range from negative 7 to 7 in one-unit increments. The line begins at point E in quadrant 2, crosses the x-axis, continues downward through quadrant 3, crosses the y-axis, and ends at point F in quadrant 4. How would you set up the distance formula to find the distance between points E 𝐸 and F 𝐹 . (1 point) Responses d=(6−2)2+(1−5)2−−−−−−−−−−−−−−−√ 𝑑 = ( 6 − 2 ) 2 + ( 1 − 5 ) 2 d is equal to square root of open paren 6 minus 2 close paren squared plus open paren 1 minus 5 close paren squared end root d=(2−6)2+(5−1)2−−−−−−−−−−−−−−−√ 𝑑 = ( 2 − 6 ) 2 + ( 5 − 1 ) 2 d is equal to square root of open paren 2 minus 6 close paren squared plus open paren 5 minus 1 close paren squared end root d=(−6−2)2+(1−(−5))2−−−−−−−−−−−−−−−−−−−√ 𝑑 = ( − 6 − 2 ) 2 + ( 1 − ( − 5 ) ) 2 d is equal to square root of open paren negative 6 minus 2 close paren squared plus open paren 1 minus negative 5 close paren s

Answer 1

Choose the coordinates E = (−6, 1) and F = (2, −5). Using the distance formula:

d = sqrt[(2 − (−6))^2 + (−5 − 1)^2] = sqrt[(−6 − 2)^2 + (1 − (−5))^2].

Either form is equivalent; evaluating gives d = sqrt(8^2 + (−6)^2) = sqrt(100) = 10.