Question

Find the values of y for which the distance between the points p (2,_3) and q (10,y) is10 units

Answer 1

HELLO DEAR,

GIVEN:- points of p(2,-3) and q(10,y)

And pq = 10units

By distance formula pq = $\bold{\sqrt{(2 - 10)^2 + (-3 - y)^2}}$

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

100 = 64 + 9 + y² + 6y

y² + 6y - 27 = 0

y² + 9y - 3y - 27 = 0

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

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

y = -9 , y = 3

so, the point at pq = 10units is -9or3

I HOPE IT'S HELP YOU DEAR,

THANKS

Answer 2

HEY DEAR ... ✌️

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=) Given ,
Points = P ( 2,-3 ) and Q ( 10,y ) .

Distance (d) = 10 units .

To find :- Value of y .

Acc. to question ,

Formula ,

d = √(a-c)^2 + (b-d)^2

Putting the values ,

=) 10 = √(2-10)^2 + ( -3-y)^2

By Squaring both the sides .

=) 100 = (-8)^2 + (9+y^2+6y)

=) 100 = 64 + 9 + y^2 + 6y

=) y^2 + 6y - 27 = 0

Solving the equation by middle term splitting.

=) y^2 + 9y - 3y - 27 =0

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

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

=) y = 3 or y = -9

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HOPE , IT HELPS ... ✌️