Math, asked by barnalipramanick90, 1 month ago

16. Find the value of y, if the distance between points (2,-3) & (10,y) is 10 units.


Answers

Answered by ItzWhiteStorm
46

Solution:

  • Here,The distance between the points (2,-3) and (10,y) is 10 units.so that we can use the formula of distance between two points to find the value of y.

\\ \underline{\dag\;\frak{As\;we\;know\;that:}} \\  \\  \green{ \bigstar} \: \fbox{\sf{Distance\;between\;points\;=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}} \\  \\

Applying the values,

\\   \implies\sf{10 =  \sqrt{{\bigg(10 - 2 \bigg)}^{2} +  {\bigg(y - ( - 3) \bigg)}^{2} }} \\  \\ \implies\sf{10 =  \sqrt{{\bigg(10 - 2 \bigg)}^{2} +  {\bigg(y + 3 \bigg)}^{2}}} \\  \\ \implies\sf{10 =  \sqrt{{ \bigg(8 \bigg)}^{2} + { \bigg(y + 3 \bigg)}^{2} }} \\  \\ \implies\sf{{10}^{2}  = 64 +  {(y + 3)}^{2}} \\  \\ \implies\sf{100 = 64 +  {(y + 3)}^{2}} \\  \\ \implies\sf{100 - 64 ={(y + 3)}^{2}} \\  \\ \implies\sf{36 =  {(y + 3)}^{2}} \\  \\ \implies\sf{ \sqrt{36} = (y + 3)} \\  \\ \implies\sf{ \pm \: 6 =y + 3} \\  \\ \implies\sf{y =  \pm \:  6 + 3} \\  \\ \implies  \underline{\boxed{\frak{y =  - 9(or)3}}} \:  \red{ \bigstar} \:  \\  \\

Hence,

  • The value of y is -9(or)3.

Answered by Anonymous
15

Given :-

  • Coordinates of A = ( 2 , - 3 )
  • Coordinates of B = ( 10 , y )
  • Distance of AB = 10 units

To Find :-

  • Value of y

Solution :-

By using distance formula

\large\red \bigstar \: \boxed{ \green{\bf  Distance =  \sqrt{ {(x_2 - x_1)}^{2} +  {(y_2 - y_1)}^{2}  } }} \\

Substitute values in formula

 \longmapsto \sf{10 = \sqrt{{\bigg(10 - 2 \bigg)}^{2} + {\bigg(y - ( - 3) \bigg)}^{2} }} \\

 \longmapsto\sf{10 = \sqrt{{\bigg(10 - 2 \bigg)}^{2} + {\bigg(y + 3 \bigg)}^{2}}} \\

 \longmapsto\sf{10 = \sqrt{{ \bigg(8 \bigg)}^{2} + { \bigg(y + 3 \bigg)}^{2} }} \\

 \longmapsto\sf{{10}^{2} = 64 + {(y + 3)}^{2}} \\

 \longmapsto\sf{100 = 64 + {(y + 3)}^{2}} \\

 \longmapsto\sf{100 - 64 ={(y + 3)}^{2}} \\

 \longmapsto\sf{36 = {(y + 3)}^{2}} \\

 \longmapsto\sf{ \sqrt{36} = (y + 3)} \\

 \longmapsto\sf{ \pm \: 6 =y + 3} \\

 \longmapsto\sf{y = \pm \: 6 + 3} \\

 \longmapsto \underline{\boxed{\frak{y = - 9(or)3}}} \: \red{ \bigstar} \: \\

Hence,

The value of y is -9 , 3

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