Question

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QUESTION 3
If the distance between the two points D(-7; -2) and F(x; 2) on the Cartesian plane
is 5 units, then calculate the possible values of x.

Answer 1

Explanation:

the distance between 2 points on a coordinate system creates a right-angled triangle.

the distance is the Hypotenuse (the side opposite of the 90° angle). and the x- and y- coordinate differences are the legs.

then Pythagoras applies

c² = a² + b²

c being the Hypotenuse, a and b being the legs.

so,

distance² = (xd - xf)² + (yd - yf)²

5² = (-7 - x)² + (-2 - 2)² = 49 + 14x + x² + 16

25 = x² + 14x + 65

x² + 14x + 40 = 0

the general solution to such a quadratic equation

ax² + bx + c = 0

is

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

x = (-14 ± sqrt(14² - 4×1×40))/(2×1) =

= (-14 ± sqrt(196 - 160))/2 =

= (-14 ± sqrt(36))/2 = (-14 ± 6)/2 = -7 ± 3

x1 = -7 + 3 = -4

x2 = -7 - 3 = -10