Verified Answer :

dalendrk

\(The\ distance\ between\ A(x_A;\ y_A)\ and\ B(x_B;\ y_B):\\\\d=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}\\\\------------------------------\\\\A(1;\ y);\ B(-2;-3);\ d=5\\\\subtitute\\\\\sqrt{(-2-1)^2+(-3-y)^2}=5\\\sqrt{(-3)^2+(-3-y)^2}=5\\\sqrt{9+(-3-y)^2}=5\Rightarrow9+(-3-y)^2=5^2\\9+(-3-y)^2=25\ \ \ \ |subtract\ 9\ from\ both\ sides\\(-3-y)^2=16\iff-3-y=-\sqrt{16}\ or\ -3-y=\sqrt{16}\\-3-y=-4\ or\ -3-y=4\ \ \ \ |add\ 3\ to\ both\ sides\\-y=-1\ or\ -y=7\ \ \ \ |change\ the\ signs\\y=1\ or\ y=-7\)

\(Answer:\boxed{(1;\ 1)\ and\ (1;-7).}\)

Cricetus

The points will be "(1, 1) and (1, -7)". A further solution is provided below.

According to the given question,

A (x₁, y₁) =  (1, y)

B (x₂, y₂) = (-2, -3)

Distance, d = 5

By using the distance formula, we get

→ \(d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

By substituting the values, we get

→ \(5 = \sqrt{(-2-1)^2+(-3-y)^2}\)

→ \(5 = \sqrt{(-3)^2+(-3-y)^2}\)

→ \(25 = 9+(-3-y)^2\)

By subtracting "9" from both sides, we get

→ \(25 -9= 9+(-3-y)^2-9\)

→        \(16=(-3-y)^2\)

→         \(\pm 4 = -3-y\)

when,

→              \(4 = -3-y\)

                \(7=-y\)

                \(y = -7\)

or,

→             \(-4=-3-y\)

               \(-1=-y\)

                 \(y = 1\)

Thus the above answer is correct.      

Learn more about distance here:

https://brainly.com/question/4058851

Related Questions