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
1.8k views
in Straight Lines by (44.3k points)

Find the point on the x-axis which is equidistant from the points (2, -5) and (-2, 9).

1 Answer

+2 votes
by (58.8k points)
selected by
 
Best answer

Let point P(x, 0) is on x-axis and equidistant from A(2, -5) and B (-2, 9)

PA = PB

or PA2 = PB2

(2 – x)2 + (– 5 – 0)2 = (– 2 – x)2 + (9 – 0)2

(2 – x)2 + (– 5)2 = (– 2 – x)2 + (9)2

29 + x2 – 4x = 85 + x2 + 4x

56 = – 8x

or x = – 7

The point on ×-axis is (– 7, 0)

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

...