Question

2. Distance between two points( x, 7) and (1, 15) is 10 units. Find the value of x.​

Answer 1

Question:

Distance between two points( x, 7) and (1, 15) is 10 units. Find the value of x.

Solution:

Given: (x,7) and (1,15) = 10 units

Need to find: Find the value of x.

Let's do it

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As we know that,

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

❍Let us consider that, $\sf{x_1 = x ,y_1 = 7 }$ and $\sf{x_2 = 1 ,y_2 = 15 }$

Applying the values,

$\to \sf{10 = \sqrt{{(1 - x)}^{2} + {(15 -7 )}^{2}}}$

$\to \sf{10 = \sqrt{ {(1 - x)}^{2} + {(8)}^{2} }}$

$\to \sf{10 = \sqrt{ {(1 - x)}^{2} + 64} } \\ \\ \to \sf{ {10}^{2} = {(1 - x)}^{2} + 64 } \\ \\ \to \sf{100 = {(1 - x)}^{2} + 64} \\ \\ \to \sf{100 - 64 = {(1 - x)}^{2} } \\ \\ \to \sf{36 = {(1 - x)}^{2} } \\ \\ \to \sf{ \sqrt{36} = (1 - x)} \\ \\ \to \sf{ \pm6 = (1 - x)} \\ \\ \to \sf{ \pm6 - 1 = x} \\ \\ \to \sf{x = 7 (or) - 5}$

• Therefore,The value of x is 7 (or)-5.

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Answer 2

Question:

• Distance between two points( x, 7) and (1, 15) is 10 units. Find the value of x.

Solution:

Given: (x,7) and (1,15) = 10 units

Need to find: Find the value of x.

Let's do it

_______________________

As we know that,

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

❍Let us consider that, $\sf{x_1 = x ,y_1 = 7 }$ and $\sf{x_2 = 1 ,y_2 = 15 }$

Applying the values,

$\to \sf{10 = \sqrt{{(1 - x)}^{2} + {(15 -7 )}^{2}}}$

$\to \sf{10 = \sqrt{ {(1 - x)}^{2} + {(8)}^{2} }}$

$\to \sf{10 = \sqrt{ {(1 - x)}^{2} + 64} } \\ \\ \to \sf{ {10}^{2} = {(1 - x)}^{2} + 64 } \\ \\ \to \sf{100 = {(1 - x)}^{2} + 64} \\ \\ \to \sf{100 - 64 = {(1 - x)}^{2} } \\ \\ \to \sf{36 = {(1 - x)}^{2} } \\ \\ \to \sf{ \sqrt{36} = (1 - x)} \\ \\ \to \sf{ \pm6 = (1 - x)} \\ \\ \to \sf{ \pm6 - 1 = x} \\ \\ \to \sf{x = 7 (or) - 5}$

Therefore,The value of x is 7 (or)-5.

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