Find the mirror image of A(-4,2) in (i) the x-axis (ii) the y-axis (iii) the origin Give the name of the figure formed by point A and the points obtained in (i) , (ii) and (iii) above. Also find the distance between poitns A and the point obtained in (iii).

Find the mirror image of A(-4,2) in (i) the x-axis (ii) the y-axis

1.	Find the mirror image of A(-4,2) in (i) the x-axis (ii) the y-axis (iii) the origin Give the name of the figure formed by point A and the points obtained in (i) , (ii) and (iii) above. Also find the distance between poitns A and the point obtained in (iii).
Answer» Solution :(-4,2) lies in II quadrant i.e, 2 units above the x-axis. (i) `THEREFORE` image B of a point A(-4,2) in x-axis will be just below the x-axis i.e. in III quadrant. So, image B`=(-4,-2)`. Since, (-4,2) lies in II quadrant i.e., 4 units left to the y-axis. `therefore` Image C of a point A(-4,2) in y-axis will be just right to the y-axis i.e, in I quadrant. so, image `C=(4,2)`. If (x,y) be any point then itsimage in the origin will be `(-x,-y)` i.e., image will be in the opposite quadrant. Here, (-4,2) lies in II quadrant, so its image D will be in IV quadrant. `therefore` Image `D= (4,-2)` Since `AB=CD=2+2=4` units and `AC=BD=4+4=8` units `therefore square ACDB` is a RECTANGLE whose LENGTH `AC=8` units and breadth `CD=4` units In right triangle `ACD`. Now, by Pythagoras theorem `AD^(2)=AC^(2)=CD^(2)` `=(8)^(2)+(4)^(2)` `=64+16` `=80` `therefore AD =sqrt(80)`units `rArrAD=4sqrt(5)` units