Question

The middle point of the line segment joining (3, -1)and (1, 1) is shifted by two units (in the sense of increasing y) Perpendicular to the line segment. Then, the coordinates of the point in the new position are

• A. $(2-\sqrt{2},2)$
• B. $(2,2-\sqrt{3})$
• C. $(2+\sqrt{2},\sqrt{2})$
• D. $(2+2,\sqrt{3})$

Answer 1

The correct option is C

$(2+\sqrt{2},\sqrt{2})$

Let P be the middle point of the line segment joining A (3, -1) and B (1, 1).

$Then,P=(\frac{3+1}{2},\frac{-1+1}{2})=(2,0)$

Let P be shifted to Q, where PQ = 2 and y – coordinate of Q is greater than that of P (from question).

$Now{,}^{\prime }slop{e}^{\prime }ofAB=\frac{1-(-1)}{1-3}=\frac{2}{2}=1{\therefore }^{\prime }slop{e}^{\prime }ofPQ=-\frac{1}{1}=-1$

$CoordinatesofQbydistanceformulaare$ $(2\pm 2cos\theta ,0,\pm 2sin\theta )$,where $tan\theta =1$

$=(2\pm 2.\frac{1}{\sqrt{2}},0\pm 2.\frac{1}{\sqrt{2}})=(2\pm \sqrt{2},\pm \sqrt{2})$

$Asy-coordinateofQ,isgreaterthanthatofP.$

$\therefore Q=(2+\sqrt{2},\sqrt{2}).Thisistherequiredpoint.$

$Hence,\left(c\right)isthecorrectanswer.$