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Transformation - class-X
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The line $3x-4y+7=0$ is rotated through an angle $\dfrac {\pi}{4}$ in clockwise direction about the point $\left (1,1\right)$. The equation of the line in its new position is
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A
$7y+x-6=0$
π‘ Explanation:
Given slope $=\dfrac{3}{4} < 1$
$< 45^{o}$
$\therefore $ after rational clock slope because negative
$\dfrac{m _1-m _{2}}{1+m _{1},m _{2}}=\tan 45$
$m _{1}=\dfrac{3}{4} \,\,\,m _{2}=$ new slope
$\left| \dfrac{\dfrac{3}{4}-m}{1+\dfrac{3}{4} m} \right|=1$
$\dfrac{3}{4}-m= \pm \left( 1+\dfrac{3}{4}m \right)$
$\dfrac{3}{4}-m=1+ \dfrac{3}{4} m$
$\dfrac{7}{4}m=\dfrac{-1}{4}$
$m=\dfrac{-1}{7}$ appeared
$\therefore$ satisfy slope and pt in option to save time.