Question

Find the value(s) of x, if the distance between the points (x, -1) and (3, 2) is 5

Answer 1

Given:

• Points are : ( x , -1 ) and ( 3 , 2)

• Distance between points is 5 unit.

To find :

• The value of x .

Solution :

• As we know that :

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

Here ,

• x1 = x and y1 = -1

• x2 = 3 and y2 = 2

• d = 5 .

$\implies \: 5 = \sqrt{ {(3 - x)}^{2} + {(2 - ( - 1))}^{2} } \\ \\ \implies \: 5 = \sqrt{ {(3 - x)}^{2} + {3}^{2} } \\ \\ \implies \: {5}^{2} = {(3 - x)}^{2} + 3 {}^{2} \\ \\ \implies \: 25 = 9 + {x}^{2} - 6x + 9 \\ \\ \implies \: {x}^{2} - 6x - 7 = 0 \\ \\ \implies \: \: {x}^{2} - 7x + x - 7 = 0 \\ \\ \implies \: x(x - 7) + 1(x - 7) \\ \\ \implies \: (x + 1)(x - 7) = 0 \\ \\ \implies \: \: x = - 1 \: and \: \: 7$

Hence ,the value of x is 7

Answer 2

Answer:

Step-by-step explanation:

Let the points (x,-1)be (x1,y1) and (3,2) be (x2.y2) have a distance of 5 units.

using distance formulae [(x1-x2)^2 + (y1-y2)^2]^1/2 we get

[(x-3)^2 + (-1 -2)^2]^1/2=5

(x-3)^2 + (-3)^2 =5^2

(x-3)^2 + 9 = 25

(x-3)^2= 16

x-3 = -4,+4

x= -1, +7

I hope this helps.