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
20.4k views
in Coordinate Geometry by (55.4k points)
closed by

If ABCD be a rectangle and P be any point in a plane of the rectangle, then prove that PA2 + PC2 = PB2 + PD2.

1 Answer

+1 vote
by (44.5k points)
selected by
 
Best answer

[Hint: Take A as the origin and AB and AD as x and y - axis respectively. Let AB = a, AD = b]

Let us assume A as the origin (0, 0) and AB and AD as x and y axis with length a and b units.

Then we get points B to be (a, 0), D to be (0, b) and C to be (a, b).

Let us assume P(x, y) be any point in a plane of the rectangle.

We need to prove PA2 + PC2 = PB2 + PD2.

We know that distance between two points (x1, y1) and (x2, y2) is

Let us assume L.H.S,

⇒ PA2 + PC2 = ((x - 0)2 + (y - 0)2) + ((x - a)2 + (y - b)2)

⇒ PA2 + PC2 = x2 + y2 + x2 - 2ax + a2 + y2 - 2by + b2

⇒ PA2 + PC2 = (x2 - 2ax + a2 + y2) + (x2 + y2 - 2by + b2)

⇒ PA2 + PC2 = ((x - a)2 + (y - 0)2) + ((x - 0)2 + (y - b)2)

⇒ PA2 + PC2 = PB2 + PD2

⇒ L.H.S = R.H.S

∴ Thus proved.

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

...