Calculate point, given angle, distance, point

Question

Given:

• Point $A(2,3)$;
• angle $45^\circ$;
• distance $AB=11$;

How to find second point?

Answer 1

The slope of the line is $\tan(45^{o} ) = 1$

So, $$\frac{y'-3}{x'-2} = 1 \implies y'-3 = x'-2 \implies y' = x'+1$$

$$AB^2 = (x'-2)^2+(y'-3)^2 = 121 \implies2(x'-2)^2 = 121 \implies x'-2 =\frac{11}{\sqrt2}$$ [$+$ because in the first quadrant, $x$ and $y$ are positive]

So, $$x' = 2+\frac{11}{\sqrt2}$$

$$y' = x'+1 = 3+\frac{11}{\sqrt2}$$

Answer 2

At $45^\circ$ you have $x_B-x_A=y_B-y_A=11\cos 45^\circ=11\frac{\sqrt 2} 2$

Answer 3

You have given $$AB=\sqrt{(x_B-2)^2+(y_B-3)^2}=11$$ and $$\tan(45^{\circ})=\frac{y_B-3}{x_B-2}$$ We get since $$\tan(45^{\circ})=1$$ so $$x_B+1=y_B$$ So we get (after squaring) $$2x_B^2-8x_B+8=121$$