Question

Distance between (7,0)and (4,k)is 5 units then k =

Answer 1

Answer:

k = +2 OR -2

Step-by-step explanation:

A = (7, 0) and B = (4,  k)

using distance formula

$AB^2= (x_{2} - x_{1} )^{2} + (y_{2} - y_{1} )^{2} \\\\5^2= (4-7)^{2} + (k-0)^2\\\\25 = (-3)^{2} + (k)^2\\\\25 = 9 + k^2\\\\K^2 = 25 - 9 = 16\\\\K^2 = (2)^2 OR (-2)^2\\\\k = -2 OR +2$

Answer 2

The value of k = - 4 , 4

Given :

Distance between (7,0) and (4,k) is 5 units

To find :

The value of k

Formula :

For the given two points ( x₁ , y₁) & (x₂ , y₂) the distance between the points

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

Solution :

Step 1 of 3 :

Find the distance between given points

Here the given points are (7,0) and (4,k)

The distance between the points

$\sf = \sqrt{ {(7 - 4)}^{2} + {(0 - k)}^{2} } \: \: unit$

$\sf = \sqrt{ {3}^{2} + {k}^{2} } \: \: unit$

$\sf = \sqrt{ 9+ {k}^{2} } \: \: unit$

Step 2 of 3 :

Form the equation

Here it is given that distance between (7,0) and (4,k) is 5 units

By the given condition

$\sf \sqrt{9 + {k}^{2} } = 5$

Step 3 of 3 :

Find the value of k

$\displaystyle \sf{ }\sqrt{9 + {k}^{2} } = 5$

$\displaystyle \sf{ \implies 9 + {k}^{2} = {5}^{2} }$

$\displaystyle \sf{ \implies 9 + {k}^{2} = 25 }$

$\displaystyle \sf{ \implies {k}^{2} = 25 - 9 }$

$\displaystyle \sf{ \implies {k}^{2} = 16 }$

$\displaystyle \sf{ \implies k = \pm \: \sqrt{16} }$

$\displaystyle \sf{ \implies k = \pm \: 4}$

Hence the required value of k = - 4 , 4

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