Question

Find the value of X if distance between two points (4, 0)and (0 x) is 5 units

Answer 1

Answer:

√(x2 - x1)²+(y2 -y1)² =5

√(0 -4)²+(x -0)²=5

16 +x² =25

x² =25 - 16

x²=9

x = 3

Step-by-step explanation:

Answer 2

The value of x is 3 or -3.

Step-by-step explanation:

Given : Distance between two points (4,0) and (0,x) is 5 units.

To find : The value of x ?

Solution :

The distance between two point $(x_1,y_1), (x_2,y_2)$ is given by,

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Here, d=5, $(x_1,y_1)=(4,0)$ and $(x_2,y_2)=(0,x)$

Substitute the value,

$5=\sqrt{(0-4)^2+(x-0)^2}$

Squaring both side,

$25=16+x^2$

$x^2=25-16$

$x^2=9$

$x=\sqrt{9}$

$x=\pm 3$

Therefore, the value of x is 3 or -3.

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