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

we know distance= sqrt[(x2-x1)^2 + (y2-y1)^2]
therefore,
10=sqrt[(10-2)^2 + (y+3)^2]
100=8^2 + y^2 + 9 +6y
100-64-9=y^2 + 6y
y^2 + 6y - 27 = 0
y^2 + 9y -3y -27 = 0
y(y+9) -3(y+9) = 0
(y+9)(y-3)=0

therefore y= -9 or y=3

Answer 2

Step-by-step explanation:

AnswEr

The Distance between the points P(2,-3) and Q(10,Y) is 10 units.

We've to find out the value of y.

Using Distance Formula,

$\dag \ \boxed{\sf{\pink{Distance = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}}}}$

Where,

$\sf{Values}\begin{cases}\sf{x_1 = 2}\\\sf{x_2 = 10}\\\sf{y_1 = -3}\\\sf{y_2 = \ 'y'}\end{cases}$

⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━⠀⠀

⠀

• Substituting values :

$:\implies\sf\purple{\sqrt{\Big( x_2 - x_1 \Big)^2 + \Big(y_2 - y_1 \Big)^2}} \\\\\\:\implies\sf \sqrt{\Big( 10 - 2 \Big)^2 + \Big( y + 3 \Big)^2} = 10 \\\\\\:\implies\sf \sqrt{\Big( 8 \Big)^2 + \Big( y + 3 \Big)^2} = 10$

• Squaring Both sides

$:\implies\sf 64 + \Big( y + 3 \Big)^2 = 100 \\\\\\:\implies\sf \Big(y + 3 \Big)^2 = 100 - 64 \\\\\\:\implies\sf \Big( y + 3 \Big)^2 = 36 \\\\\\:\implies\sf\pink{ y + 3 = \pm 6} \\\\\\:\implies\sf y + 3 = 6\\\\\\:\implies\sf y = 6 - 3\\\\\\:\implies\boxed{\frak{\pink{ y = 3}}}\\\\\\:\implies\sf y + 3 = - 6\\\\\\:\implies\sf y = - 6 - 3\\\\\\:\implies\boxed{\frak{\pink{y = -9}}}$

$\therefore\underline{\textsf{ Hence, value of y is \textbf{3 or -9}}}.$