Question

1. The distance between the points (2,-3) and (K,9) is 13 units.Find 'k'.​

Answer 1

☯ AnSwEr :

We know the distanve formula.

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

Where,

• x1 = 2
• x2 = K
• y1 = -3
• y2 = 9
• Distance = 13 units

★ Putting Values ★

$\sf{\dashrightarrow 13 = \sqrt{(k - 2)^2 + \bigg(9 - (-3) \bigg)^2}} \\ \\ \sf{\dashrightarrow (13)^2 = k^2 + (k - 2)^2 + (12)^2} \\ \\ \sf{\dashrightarrow 169 = (k - 2)^2 + 144} \\ \\ \sf{\dashrightarrow (k - 2)^2 = 169 - 144} \\ \\ \sf{\dashrightarrow (k - 2)^2 = (5)^2} \\ \\ \sf{\dashrightarrow k - 2 = 5} \\ \\ \sf{\dashrightarrow k = 5 + 2} \\ \\ \sf{\dashrightarrow k = 7} \\ \\ \Large{\implies{\boxed{\boxed{\sf{k = 7}}}}}$

Answer 2

★ Answer :

We know that,

$\Large{\star{\boxed{\tt{Distance = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} }}}}$

Where,

• $\tt{x_1 = 2}$
• $\tt{x_2 = k}$
• $\tt{y_1 = -3}$
• $\tt{y_1 = 9}$

$\tt{\implies 13 = \sqrt{(k - 2)^2 + \big(9 - (-3) \big)^2}} \\ \\ \tt{\implies (13)^2 = k^2 + (k - 2)^2 + (12)^2} \\ \\ \tt{\implies 169 = (k - 2)^2 + 144} \\ \\ \tt{\implies (k - 2)^2 = 169 - 144} \\ \\ \tt{\implies (k - 2)^2 = 25} \\ \\ \tt{\implies (k - 2)^2 = (5)^2} \\ \\ \tt{\implies k - 2 = 5} \\ \\ \tt{\implies k = 5 + 2} \\ \\ \tt{\implies k = 7} \\ \\ \Large{\star{\boxed{\tt{k = 7}}}}$

So, value of k is 7.