Question

If the distance between point (x, 0) and (0,3) is 5, what are value of x

Answer 1

(Dis)² = (x2 - x1)² +(y2 - y1)²
25 = ( 0 - x)² + ( 3 -0)²
25 = (-x)² + (3)²
25 = x² + 9
16 = x²
x = 4

Answer 2

Answer:

4 or - 4

Step-by-step explanation:

Formula:

$\text {Distance = } \sqrt{ (Y_1 - Y_2)^2 + (X_2 - X_1)^2}$

Find the distance in term of x:

$\text {Distance = } \sqrt{ (0 - 3)^2 + (x - 0)^2}$

$\text {Distance = } \sqrt{ (- 3)^2 + (x)^2}$

$\text {Distance = } \sqrt{ 9 + x^2}$

Solve x:

Given that the distance is 5

$\sqrt{ 9 + x^2}  = 5$

$(\sqrt{ 9 + x^2} )^2 = 5^2$

$9 + x^2 = 25$

$x^2 = 16$

$x = \pm 4$

Answer: The value of x can be 4 or - 4.