PQ is a straight line of 13 units if P has coordinates (5,-3) and Q has coordinates (-7,y) find the value of y

2 thoughts on “PQ is a straight line of 13 units if P has coordinates (5,-3) and Q has coordinates (-7,y) find the value of y”

1. Answer:

   by distance formula

   y-3/-7+5=3

   y-3=13(-2)

   y=3-26

   y=-23

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2. Given :

   •PQ is a straight line

   •Distance between PQ=13units

   •Coordinates of P=(5,-3)

   • Coordinates of Q=(-7, y)

   To Find :

   • The value of y=?

   Formula :

   Distance formula :

   [tex]\boxed{\bf{ \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}}}[/tex]

   Solution :

   Given Points :

   •P=(5,-3)

   •Q=(-7,y)

   By using distance formula :

   [tex]\implies\sf{\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}}[/tex]

   [tex] \\ \implies \sf \: \sqrt{ {(y – ( – 3)}^{2} + {( – 7 – 5)}^{2} } [/tex]

   [tex] \\ \implies \sf \: \sqrt{ {(y + 3)}^{2} + 144} = distance = 13[/tex]

   [tex] \\ \implies \sf \: {(y + 3)}^{2} = 169 – 144[/tex]

   [tex] \\ \implies \sf \: {y}^{2} + 9 + 6y = 25[/tex]

   [tex] \\ \implies \sf \: {y}^{2} + 6y = 25 – 9[/tex]

   [tex] \\ \implies \sf \: {y}^{2} + 6y – 16 = 0[/tex]

   [tex] \\ \implies \sf \: {y}^{2} + 8y – 2y – 16 = 0[/tex]

   [tex] \\ \implies \sf \: y(y + 8) – 2(y + 8) = 0[/tex]

   [tex] \\ \implies \sf \boxed{ \bf{y = – 8} \ } \: and \: \boxed{ \bf{y = 2}}[/tex]

   Therefore, The value of y is 2 .

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