Question

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

Answer 1

Answer:

Given

P(2,-3)

Q(10,y)

distance PQ = 10 units

to find

value of y

Solution

X1 = 2

X2 = 10

Y1 = -3

Y2 = y

so distance PQ

________________

=>√ (X2 - X1)² + (Y2 - Y1)² = 10

____________

=> √(10-2)²+ {y-(-3)}² = 10

________

=> √ 8²+(y+3)² = 10

=> 64 + y²+6y + 9 = 10².

[ taking square root to rhs to become square ]

=> y²+6y = 100-64-9

=> y²+6y = 27

=> y²+6y-27 = 0

=>y²+9y-3y -27 =0

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

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

=> y = -9 , 3

Step-by-step explanation:

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