Question

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

Answer 1

Answer:

by distance formula

y-3/-7+5=3

y-3=13(-2)

y=3-26

y=-23

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Answer 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 :

$\boxed{\bf{ \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}}}$

Solution :

Given Points :

•P=(5,-3)

•Q=(-7,y)

By using distance formula :

$\implies\sf{\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}}$

$\\ \implies \sf \: \sqrt{ {(y - ( - 3)}^{2} + {( - 7 - 5)}^{2} }$

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

$\\ \implies \sf \: {(y + 3)}^{2} = 169 - 144$

$\\ \implies \sf \: {y}^{2} + 9 + 6y = 25$

$\\ \implies \sf \: {y}^{2} + 6y = 25 - 9$

$\\ \implies \sf \: {y}^{2} + 6y - 16 = 0$

$\\ \implies \sf \: {y}^{2} + 8y - 2y - 16 = 0$

$\\ \implies \sf \: y(y + 8) - 2(y + 8) = 0$

$\\ \implies \sf \boxed{ \bf{y = - 8} \ } \: and \: \boxed{ \bf{y = 2}}$

Therefore, The value of y is 2 .