The line $2x-y=5$ turns about a point on it, whose ordinate and abscissae are equal, through an angle of $45°$, in anti clockwise direction. Find the equation of line in the new position.
My attempt to solve: Let AB be the line with the equation $2x-y=5$ and $B(a,a)$ be a point on it whose ordinate and abscissae are equal. Then $$2x-y=5$$ $$2a-a=5$$ $$a=5$$ Hence, $B(a,a)=B(5,5)$.
My solution ends up here. What should I do next?
On
hint use rotation matrix which goes as $$\left(\begin{matrix} \cos45&\sin45\\ -\sin45 & \cos 45\end {matrix} \right)$$ and represent line as a form of matrix multiply these two matrices and get the equation .
On
The line turns by an angle of $45°$ anti-clockwise about the point $(5,5)$. Thus, you need to find the equation of a line whose angle with the $x-axis$ is $45°$ more than this line and which passes through the point $(5,5)$.
The line $2x-y=5$ has a slope of $2$, i.e., it makes an angle of $\arctan2$ with the $x-axis$.
Slope of the required line=$\tan(\frac\pi4+\arctan2)=\frac{1+2}{1-2}=-3$
Its equation, therefore, is $$y-5=-3(x-5)$$ or $$3x+y=20$$
In new position the line will pass through the point of rotation $(5, 5)$ & the slope of the new line becomes $m=\tan\left(\tan^{-1}(2)+\frac{\pi}{4}\right)=-3$ hence the new equation of the line, using point-slope formula: $y-y_1=m(x-x_1)$,
$$y-5=\left(-3\right)(x-5)$$ $$3x+y=20$$