Question

P(5,-3) and Q(-7,y) and PQ=13 units, find y​

Answer 1

$\text{Formula used:}$

$\text{The distance between two points (x_1,y_1) and (x_2,y_2) is}$

$\boxed{\bf\:d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}}$

$\text{Given points are}$

$\text{P(5,-3) and Q(-7,y)}$

$\text{Now, PQ=13}$

$\implies\,\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}=13$

$\implies\,\sqrt{(5+7)^2+(-3-y)^2}=13$

$\implies\,\sqrt{144+(3+y)^2}=13$

$\text{Squaring on both sides, we get}$

$\implies\,144+(3+y)^2=169$

$\implies\,(3+y)^2=25$

$\implies\,3+y=\pm\,5$

$\implies\,y=-3\pm\,5$

$\implies\,y=-3+5,\;-3-5$

$\implies\,y=2,\;-8$

$\therefore\textbf{The values of y are 2 and -8}$

Answer 2

y=-8 or y= 2

•P(5,-3) and Q(-7,y)

• By distance formula

PQ = √[(X2-X1)²+(Y2-Y1)²]

PQ² = (X2-X1)²+(Y2-Y1)²

13² = (-7-5)² + (y+3)²

169 = (12)² + y² +9 +6y

0 = y² + 6y -16

0 = y² +8y -2y -16

0 = y( y+8 )-2( y+8 )

0 = ( y+8 )( y-2 )

y=-8 or y= 2