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

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”

by distance formula

y-3/-7+5=3

y-3=13(-2)

y=3-26

y=-23

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 .