wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

AB is a line segment which is 8 cm long. CD and EF are parallel to AB, at a distance of 4 cm from it. GH is a line segment parallel to AB, at a distance of 2 cm from it. PQ is the perpendicular bisector of AB which intersects CD, GH, AB and EF at M, N, O and P respectively.

The point/s which is/are always at a distance of 4 cm from the line AB and equidistant from A and B is/are


Open in App
Solution

We know that the locus of a point which is at a given distance from a given line is a pair of lines parallel to the given line and at a given distance from it.

Hence in the given figure, the locus of the point which is always at a distance of 4 cm from the line AB are the lines CD and EF.

Also, we know that the locus of a point which is equidistant from two fixed points is the perpendicular bisector of the line segment joining the two fixed points.

Hence in the given figure, the locus of the point which is equidistant from A and B the perpendicular bisector of the line segment joining A and B, i.e. the perpendicular bisector PQ.

Hence the points which satisfy both the conditions, i.e. always at a distance of 4 cm from the line AB and equidistant from A and B are the point of intersection of CD and PQ, which is M and the point of intersection of EF and PQ, which is P.


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
perpendicular bisector to a given line segment
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon