Question

determine o valor real de k de modo que a distância entre os pontos a(k-3, 2) e b(2k-1, -2) seja igual a 5

Answer 1

dAB = $\sqrt{(xB-xA^2+(yB-yA)^2)}$

xA = k - 3
yA = 2

xB = 2k - 1
yB = -2

5 = $\sqrt{[2k-1-(k-3)]^2+(-2-2)^2}$

5 = $\sqrt{(2k-1-k+3)^2+ (-4)^2}$

5 = $\sqrt{(k+2)^2+16}$

5² = ($\sqrt{(k+2)^2+16}$)²

25 = k² + 4k + 4 + 16

25 = k² + 4k + 20

k² + 4k - 5 = 0

a = 1
b = 4
c = -5

Δ = b² - 4ac
Δ = (4)² - 4.(1).(-5)
Δ = 16 + 20
Δ = 36

k = $\frac{-b \frac{+}{-} \sqrt{b^2-4ac} }{2a}$

k = $\frac{-4 \frac{+}{-} \sqrt{36} }{2}$

k = $\frac{-4 \frac{+}{-} 6}{2}$

k' = (- 4 + 6) / 2

k' = 2 / 2

k' = 1

k" = (-4 - 6) / 2

k" = -10/2

k" = -5

k = 1 ou k = -5

S = {-5,1}