Point P (2$$, $$4) is translated through a $$distance32$$ units measured parallel to the line y $$-$$ x $$-1$$ $$=$$ $$0$$, in the direction of decreasing ordinates, to reach at Q. If R is the image ofQ unth respect to the line y $$-$$ x $$-1$$ $$=$$ $$0$$, then coordinates of R are

Question

Point P (2$$, $$4) is translated through a $$distance32$$ units measured parallel to the line y $$-$$ x $$-1$$ $$=$$ $$0$$, in the direction of decreasing ordinates, to reach at Q. If R is the image ofQ unth respect to the line y $$-$$ x $$-1$$ $$=$$ $$0$$, then coordinates of R are

Infinity Learn · Accepted Answer

The equation of a line through P (2$$, $$4) and parallel to y $$-$$ x $$-$$ $$1$$ $$=$$ $$0$$ isx−$$2cos$$⁡π$$/4=y$$−$$4sin$$⁡π$$/4The$$ coordinates of Q are given byx−$$2cos$$⁡π$$/4=y$$−$$4sin$$⁡π$$/4=$$−$$32$$⇒$$x=$$−$$1$$,$$y=1Thus$$, the coordinates of Q are $$(-1$$, $$1)$$. R s the image of Q with respect to the line y $$-$$ x $$-1$$ $$=$$ $$0$$.Therefore, coordinates of R are given $$byx+1$$−$$1=y$$−$$11=$$−$$2(1+1$$−$$1)($$−$$1)2+(1)2x+1$$−$$1=y$$−$$11=$$−$$1$$⇒$$x=0$$,$$y=0Hence$$, the coordinates of R are (0$$, $$0).

Point P $(2, 4)$ is translated through a distance $3 \sqrt{2}$ units measured parallel to the line $y - x - 1 = 0$ , in the direction of decreasing ordinates, to reach at $Q .$ $I f R$ is the image ofQ unth respect to the line $y - x - 1 = 0$ , then coordinates of $R$ are

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Point P (2, 4) is translated through a distance32 units measured parallel to the line y - x -1 = 0, in the direction of decreasing ordinates, to reach at Q. If R is the image ofQ unth respect to the line y - x -1 = 0, then coordinates of R are

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Point P $(2, 4)$ is translated through a distance $3 \sqrt{2}$ units measured parallel to the line $y - x - 1 = 0$ , in the direction of decreasing ordinates, to reach at $Q .$ $I f R$ is the image ofQ unth respect to the line $y - x - 1 = 0$ , then coordinates of $R$ are