Question

find the values of y for which the distance between the points P(2.-3) and Q(10.) is
10 units​

Answer 1

Answer:

The value of y is 5.

Step-by-step explanation:

The formula to compute distance between two points is:

$d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

The two points are: P (2, -3) and Q (10, y).

The distance PQ = 10 units.

Compute the value of y as follows:

$PQ=\sqrt{(10-2)^{2}+(y-(3))^{2}}\\10^{2}=36+(y-(3))^{2}\\(y-(3))^{2}=100-36\\(y-(3))^{2}=64\\(y-(3))=\sqrt{64}\\y+3=8\\y=8-3\\=5$

Thus, the value of y is 5.

Answer 2

Answer:

3 or -9

Step-by-step explanation:

√(10-2)² + (y+3)² = 10

√(8)² + y² + 9 +6y=10

73+y²+6y=100

y²+6y=27

y²+6y-27=0

y²+9y-3y-27=0

y(y+9)-3(y+9)=0

(y-3)(y+9)=0

y=3 or y= -9.