Question

Find x so the distance between the points (x,2) and (-1,-2) is 1" title="2\sqrt{5}" alt="2\sqrt{5}" align="absmiddle" class="latex-formula">
x =_____ , ______

Answer 1

Answer:

x = -5 , 3

Step-by-step explanation:

(x₁ , y₁) = (-1, -2)    & (x₂, y₂) = (x , 2)

$Distance =\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\\\\\\\sqrt{(x - [-1])^{2}+(2-[-2])^{2}}=2\sqrt{5}\\\\\sqrt{(x+1)^{2}+(2+2)^{2}}=2\sqrt{5}\\\\\sqrt{(x)^{2}+2x+1+(4)^{2}}=2\sqrt{5}\\\\\sqrt{x^{2}+2x+1+4}=2\sqrt{5}\\\\\sqrt{x^{2}+2x+5}=2\sqrt{5}$

Take square both sides,

x² + 2x + 5 = (2)²(√5)²               {(√5)² = √5*√5 = 5 }

x² + 2x +5 = 4 * 5

x² + 2x + 5 = 20

x² + 2x + 5 - 20 = 0

x² + 2x  -15 = 0

x² + 5x - 3x  - 5*3 = 0

x(x + 5) -3(x + 5)=0

(x + 5)(x - 3) = 0

x + 5 = 0     ; x - 3 = 0

x = -5        ;    x = 3

x = -5 , 3