Question

if the distance between (2, 3) and (8, k) is 10 find the value of k.

Answer 1

SOLUTION :

DISTANCE BETWEEN ANY TWO POINTS = √[ (x₂-x₁)² + (y₂-y₁)²]

GIVEN POINTS = (2,3) (8,k)

x₁, y₁ x₂ ,y₂

GIVEN DISTANCE = 10

=> √[ (8-2)² + ( k-3)² ] = 10

=> √[ 36 + (k-3)² ] = 10

SQUARING ON BOTH SIDES ,

=> 36 + (k-3)² = 100

=> 36 + k² + 9 - 6k = 100

{ ∵ (a-b)² = a² + b² - 2ab }

=> k² - 6k = 100 - 45

=> k² - 6k - 55 = 0

=> k² + 5k - 11k - 55 = 0

=> k( k + 5 ) -11 ( k + 5 ) = 0

=> ( k - 11 )( k + 5 ) = 0

=> k = 11 (or) k = -5

HOPE IT HELPS !!!

Answer 2

Step-by-step explanation:

Distance Formula :-

$\sqrt{ {(x2 - x1)}^{2} + {(y2 - y1)}^{2} }$

Given :-

$Distance \: between \: the \: points = 10 \\ = > \sqrt{ {(2 - 8)}^{2} + {(3 - k)}^{2} } = 10\\ = \sqrt{ { ( - 6)}^{2} + {(3)}^{2} + {k}^{2} + 2 \times 3 \times k } = 10 \\ = \sqrt{36 + 9 + {k}^{2} + 6k } = 10 \\ = > \sqrt{45 + {k}^{2} + 6k } = 10 \\ = > {k}^{2} + 6k + 45 = {10}^{2} \\ = > {k}^{2} + 6k + 45 - 100 = 0 \\ = > {k}^{2} + 6k + ( - 55) = 0 \\ = > {k}^{2} + 6k - 55 = 0 \\ = > {k}^{2} + 11k - 5k - 55 = 0 \\ = > k(k + 11) - 5(k + 11) = 0 \\ = > (k - 5)(k + 11) = 0 \\ = > k \: = 5 \: ( \: or \: ) \: - 11$

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