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 About the author Claire

Answer: by distance formula y-3/-7+5=3 y-3=13(-2) y=3-26 y=-23 mark my answer as brainlist please. Reply

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 . Reply

Answer:by distance formula

y-3/-7+5=3

y-3=13(-2)

y=3-26

y=-23

mark my answer as brainlist please.

Given:•PQ is a straight line

•Distance between PQ=13units

•Coordinates of P=(5,-3)

• Coordinates of Q=(-7, y)

ToFind:• The value of y=?

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

Solution:Given Points :

•P=(5,-3)

•Q=(-7,y)

Byusingdistanceformula:[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,

Thevalueofyis2.