Question

find the value of Y for which the distance between the points p(2,-3) and Q(10,Y) is 10 units​

Answer 1

Given:-

• Coordinates of P = (2,-3)

• Coordinates of Q = (10,Y)

• Distance b/w them = 10units.

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To Find:-

Value of Y.

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Solution :-

$\sf Distance Formula = \sqrt{(x_{2} - x_{1}) {}^{2} + (y_ {2} - y_{2}) {}^{2} }$

10 = $\sf \sqrt{[10 - 2] {}^{2} + [Y- (-3)] {}^{2} }$

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

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

10 = $\sf\sqrt{64 + y{}^{2} + 9 + 6y}$

100 = 64 + y² + 9 + 6y

y² + 6y -27 = 0

(y+9) (y-3)

$\boxed{\bf{y = -9 \: and \: 3}}$

Therefore, value of y is -9 and 3.

Answer 2

Step-by-step explanation:

Given : -

• the points p(2,-3) and Q(10,Y) is 10 units

To Find : -

• find the value of Y for which the distance

Solution : -

the formula to calculate the distance between two points (a,b) and (c,d) is given by

dist d = √(a - c)² + (b - d)²

applying it for this ques we have d=10 units

➻ 10 = √(2 - 10)² + ( - 3 - y)²

Squaring both sides :

➻ 100 = (-8)² + (9 + y² + 6y)

➻ 100 = 64 + 9 + y² + 6y

➻ y² + 6y = 100 - 73

➻ y² + 6y - 27 = 0

➻ y² + 9y - 3y - 27 = 0

➻ y (y + 9) - 3 (y + 9) = 0

➻ (y - 3) (y + 9) = 0

➻ y = 3 or y = -9

both values of y are possible unless no other condition is specified in the question.