Question

the value of x when the distance between the points (7,x) and (4,0) is 5 units is

Answer 1

Let,

x1 = 7 y1 = x

x2=4 y2= 0

By distance formula,

√[x1-x2]² + [y1-y2]² = 5 ___________[given]

√[7-4]²+[x-0]² = 5

squaring both sides,

3² + x² = 5²

9+ x² = 25

x² = 16

x = √16

x =±4

Answer 2

Answer:

The value of x = ± 4

Step-by-step explanation:

We know that,

Distance between two points $(x_{1} ,y_{1} ) and (x_{2}, y_{2} )$ =$\sqrt{(x_{2}- x_{1} )^{2} +( y_{2}- y_{1} )^{2} }$

According to the question,

Distance between the two points = 5 units.

The points $(x_{1} ,y_{1} ) and (x_{2}, y_{2} )$ are (7,x) and (4,0) respectively.

Using distance formula,

5 = $\sqrt{(4- 7 )^{2} +( 0- x)^{2} }$

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

Squaring both sides,

25 = 9+$x^{2}$

$x^{2}$ = 16

x = ± 4

The correct answer is x = ± 4