Question

Find the values of y for which the distance between the points P(2,-3) and Q(10,y) is 10 unit​

Answer 1

Given :-

• The distance between the points P(2,-3) and Q(10,y) is 10 unit.

To find :-

• Value of y

Solution :-

• Distance between P and Q point = 10unit

According to distance formula

→ PQ = √(x2 - x1)² + (y2 - y1)²

• x1 = 2 x2 = 10

• y1 = - 3 y2 = y

→ 10 = √(10 - 2)² + (y + 3)²

→ 10 = √(8)² + y² + 9 + 6y

→ 10 = √64 + y² + 9 + 6y

→ 10 = √73 + y² + 6y

Squaring both side

→ 100 = 73 + y² + 6y

→ 73 - 100 + y² + 6y = 0

→ y² + 6y - 27 = 0

→ y² + 9y - 3y - 27 = 0

→ y(y + 9) - 3(y + 9) = 0

→ (y + 9)(y - 3) = 0

Either

→ y + 9 = 0

→ y = - 9

Or

→ y - 3 = 0

→ y = 3

• Distance never be in negative

Hence,

• Value of y is 3 unit

Answer 2

The given point P ( 2 , -3 ) and Q ( 10 , y )

$PQ\:=\:√(\:10-2\:)^2\:+\:[\:4-(-3)^2\:]$

= $√8\:+\:(\:y\:+\:3\:)^2$

= $√64\:+\:y^2\:+\:6y\:+\:9$

= $√y^2\:+\:6y\:+\:73$

But PQ = 10 { given }

So,

⇛$√y^2\:+\:6y\:+\:73\:=\:10$

squaring both sides ,

⇛$y^2\:+\:6y\:+\:73\:=\:100$

⇛$y^2\:+\:6y\:-\:27\:=\:0$

⇛$y^2\:-\:3y\:+\:9y\:-\:27\:=\:0$

⇛$either\:(\:y\:-\:3\:)\:(\:y\:+\:9\:)\:=\:0$

⇛$either\:y\:-\:3\:=\:0\:⇛\:y\:=\:3$

⇛$y\:+\:9\:=\:0\:⇛\:y\:=\:-9$

$Therefore, the\:value\:of\:<strong>y\:=\:3\:or\: 9</strong>$

♥ So, It's Done ♥ !!