Answer :
[tex]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[/tex]
[tex]Answer:\boxed{(1;\ 1)\ and\ (1;-7).}[/tex]
[tex]Answer:\boxed{(1;\ 1)\ and\ (1;-7).}[/tex]
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
→ [tex]d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
By substituting the values, we get
→ [tex]5 = \sqrt{(-2-1)^2+(-3-y)^2}[/tex]
→ [tex]5 = \sqrt{(-3)^2+(-3-y)^2}[/tex]
→ [tex]25 = 9+(-3-y)^2[/tex]
By subtracting "9" from both sides, we get
→ [tex]25 -9= 9+(-3-y)^2-9[/tex]
→ [tex]16=(-3-y)^2[/tex]
→ [tex]\pm 4 = -3-y[/tex]
when,
→ [tex]4 = -3-y[/tex]
[tex]7=-y[/tex]
[tex]y = -7[/tex]
or,
→ [tex]-4=-3-y[/tex]
[tex]-1=-y[/tex]
[tex]y = 1[/tex]
Thus the above answer is correct.
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