The line joining the points A(2,0) and B(3,1) | vedclass

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Question

(A) The slope of line $AB$ is $m = \frac{1-0}{3-2} = 1$. Since $	an 	heta = 1$,the angle that $AB$ makes with the positive $x$-axis is $	heta = 45^

Vedclass · Accepted Answer

$(2, \sqrt{2})$

Vedclass · Answer

(A) The slope of line $AB$ is $m = \frac{1-0}{3-2} = 1$. Since $	an 	heta = 1$,the angle that $AB$ makes with the positive $x$-axis is $	heta = 45^{\circ}$. When the line is rotated by $45^{\circ}$ in the anti-clockwise direction about point $A$,the new angle with the positive $x$-axis becomes $45^{\circ} + 45^{\circ} = 90^{\circ}$. The length of the segment $AB$ is $r = \sqrt{(3-2)^2 + (1-0)^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$. Let the new position of $B$ be $C(x, y)$. Since the length remains constant,$AC = AB = \sqrt{2}$. The coordinates of $C$ are given by $(x_A + r \cos \phi, y_A + r \sin \phi)$,where $\phi = 90^{\circ}$. $x = 2 + \sqrt{2} \cos 90^{\circ} = 2 + \sqrt{2}(0) = 2$. $y = 0 + \sqrt{2} \sin 90^{\circ} = 0 + \sqrt{2}(1) = \sqrt{2}$. Thus,the new coordinates are $(2, \sqrt{2})$.

The line joining the points $A(2,0)$ and $B(3,1)$ is rotated through an angle of $45^{\circ}$ about $A$ in the anti-clockwise direction. Find the coordinates of $B$ in the new position.

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