Question

what are the values of y if the distance between the points (2,y) and (-4,3) is 10?​

Answer 1

Answer:

Step-by-step explanation:

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

It is given that the distance between (2, - 3) and (10, y) is 10.

Therefore, \sqrt{(2 - 10)^2 + (-3 - y)^2 = 10

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


64 + (y + 3)2 = 100

(y +3)2 = 36

y + 3 = ±6

y + 3 = +6 or y + 3 = -6

Therefore, y = 3 or -9

Hope this helps!

Answer 2

• The distance formula is an algebraic expression used to determine the distance between two points with the coordinates (x₁, y₁) and (x₂, y₂).

Distance Between Two Points is Given by : $\sf{\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}}$

• Plug In the values

$\Rightarrow \textsf{10 units} = \sf{\sqrt{\left(-4-2\right)^{2}+\left(3-y\right)^{2}}}$

$\Rightarrow \textsf{10 units} = \sf{\sqrt{\left(-6\right)^{2}+\left(3-y\right)^{2}}}$

$\Rightarrow \textsf{10 units} = \sf{\sqrt{36+\left(3-y\right)^{2}}}$

• Squaring On Both Sides

$\Rightarrow \textsf{100 units} = \sf{36+\left(3-y\right)^{2}}$

• Subtracting 36 from Both Sides

$\Rightarrow \textsf{100 - 36 units} = \sf{\left(3-y\right)^{2}}$

$\Rightarrow \textsf{64 units} = \sf{\left(3-y\right)^{2}}$

• Taking Square Root on Both Sides

$\Rightarrow \sf{\pm}\textsf{8 units} = \sf{3-y^{}}$

• Then, The Possible Solution is

⇒ + 8 = 3 - y

⇒ + 8 - 3 = - y

⇒ + 5 = - y

⇒ - 5 = y

• And Other Possible Solution is

⇒ - 8 = 3 - y

⇒ - 8 - 3 = - y

⇒ - 11 = - y

⇒ 11 =  y