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

AnswEr :

$\bf{\Large{\underline{\sf{Given\::}}}}$

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

$\bf{\Large{\underline{\sf{To\:find\::}}}}$

The value of y.

$\bf{\Large{\underline{\underline{\tt{\red{Explanation\::}}}}}}$

$\bigstar\bf{\Large{\underline{\sf{Distance\:formula\::}}}}}\\\\\bf{\large{\boxed{\sf{PQ\:=\:\sqrt{(x_{2}-x_{1})^{2} +(y_{2}-y_{1})^{2} } }}}}}$

We have,

• x1 = 2 & y1 = -3
• x2 = 10 & y2 = y

A/q

$\implies\sf{PQ\:=\:\sqrt{(10-2)^{2} +\big(y-(-3)\big)^{2} }}$

$\implies\sf{10\:=\:\sqrt{(8)^{2}+(y+3) ^{2} } }\\\\\\\implies\sf{10^{2} \:=\:(8)^{2} +(y+3)^{2} }\\\\\\\implies\sf{100\:=\:64+y^{2}+3^{2} +2*y*3}\\\\\\\implies\sf{100\:=\:64+y^{2} +9+6y}\\\\\\\implies\sf{y^{2} +6y+9+64-100=0}\\\\\\\implies\sf{y^{2} +6y+73-100=0}\\\\\\\implies\sf{y^{2} +6y-27=0}\\\\\\\implies\sf{y^{2} +9y-3y-27=0}\\\\\\\implies\sf{y(y+9)-3(y+9)=0}\\\\\\\implies\sf{(y+9)(y-3)=0}\\\\\\\implies\sf{y+9=0\:\:\:\:\:\:\:OR\:\:\:\:\:\:y-3=0}$

$\implies\sf{\red{y\:=\:-9\:\:\:\:\:\:\:\:Or\:\:\:\:\:\:\:y\:=\:3}}$

Thus,

The value of y is -9 or 3.

Answer 2

Step-by-step explanation:

Let the points be P(2,-3) and Q(10,y)

Given, PQ = 10 units

By distance formula,

PQ = √(x₂-x₁)² + (y₂-y₁)²

x₁ = 2, y₁= -3

x₂ = 10, y₂ = y

PQ = √(10-2)² + (y-(-3))²

10 = √(8)² + (y+3)²

Squaring both sides,

10² = (√(8)² + (y+3)²)²

10² = 8²+(y+3)²

100 = 64+ y² + 3² + 2×3×y

100 = 64+ y²+9+6y

0 = y² + 6y + 64 + 9 - 100

y² + 6y - 27 = 0

y² + 9y - 3y - 27 = 0

y(y+9) - 3(y+9) = 0

(y-3)(y+9) = 0

Hence, y = 3, y = -9 is the solution.