Question

Find the value of y for which the distance between the points p(2 -3) and q(10 y) is 10 units​

Answer 1

Answer:

Step-by-step explanation:

Using the distance formula: distance = $\sqrt{(x_{2} -x_{1})^{2}+(y_{2} -y_{1})^{2} }$

sub it in,

10 = $\sqrt{(10 -2)^{2}+(y-(-3)^{2}}$

10 = $\sqrt{(8)^{2} +(y+3)^{2}}$

10 = $\sqrt{64+y^{2} +9+6y}$

10 = $\sqrt{73+y^{2} +6y}$

square both sides:

100 = 73 + $y^{2}$ + 6y

$y^{2}$ + 6y = 27

y( y + 6) = 27

y = 27 OR THE OTHER SOLUTION IS y + 6 = 27

y = 21