Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
216 views
in Geometry by (71.6k points)
closed by
Show that the locous of a point equidistant from the end points a line segment is the perpendicular bisector of the segment.

1 Answer

0 votes
by (71.6k points)
selected by
 
Best answer
The proof will be taken up in two steps :
Step 1: We, initially prove tha tany point equidistant form the end point of a line segment lies on the perpendicular bisector of the line segment.
Given : M and N are two points on a plane. A is a point in the same plane such that `AM=AN`.
RTP: A lies on the perpendicular bisector of MN.
Proof : Let M and N be the two fixed points in a plane
Let A be a point such that `AM=AN` and L be the mid -points of `overline(MN)`.
If A coincides with L, then A lies on the bisector of MN. Suppose A is different from L.
Then, in `Delta MLA and Delta NLA`,
`ML=NL,AM=AN` and AL is a common side.
`therefore` By SSS congruence property, `Delta MLA approx Delta NLA`
`rArr angle MLA =angle NLA(because "Corresponding elements of congruent triangles are equal") (1)`
But `angle MLA +angle NLA =180^@(because "They form a straight angle")`
`rArr 2 angle MLA =180^@("using(1))`
`therefore angle MLA angle NLA =90^@`
So, `overline(AL) bot overline (MN)` and hence `overline(AL)` is the perpendicular bisector of `overline(MN)`.
`therefore` A lies on the perpendicular bisector of `overline(MN)`.
Step 2 : Now, we prove that any point of the line segment.
Given : MN is the segment and P is point on the perpendicular bisector. L is the mid-point of MN.
RTP : `MP=NP`
Proof : If P coincides with L, then MP=NP. Suppos P is different from L. Then, in `Delta MLP and NLP, ML=LN`
LP is the common side and `angle MLP=angle NLP=90^@`
`therefore` By the SAS congruence property , `Delta MLPapprox Delta NLP`.
So, `MP=PN` (`because` The corresponding bisector of `overline(MN)` is eqidistant from the points M and N .
Hence , from the steps 1 and 2 of the proof it can be said that the locus of the point equidistant from two fixed points is the perpendicular bisector of the line segment joining the two points.
image
image

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...