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
6.7k views
in Coordinate Geometry by (24.0k points)
closed by

If (–2, 6) is the image of the point (4, 2) with respect to the line L = 0, then L is equal to

(a) 2x + 3y – 5 

(b) 3x – 2y + 5 

(c) 4x + 6y – 5 

(d) 6x – 4y + 5

1 Answer

+1 vote
by (23.5k points)
selected by
 
Best answer

(b) 3x – 2y + 5 

Let the equation of the L be y = mx + c 

Since O′ (–2, 6) is the image of the point O (4, 2) in line 

L = 0, the mid-point of OO′, i.e., \(\bigg(\frac{-2+4}{2},\frac{6+2}{2}\bigg)\), i.e.,

(1, 4) will lie on the given line. 

Also, L ⊥ OO′, so 

Slope of L x Slope of OO′ = –1

⇒ m x \(\bigg(\frac{6-2}{-2-4}\bigg)\)= -1 ⇒ m = \(\frac{-1}{\frac{-2}{3}}\) = \(\frac{3}{2}\)

∴ Equation of L is y = \(\frac{3}{2}\,x+c\)

∵ It passes through (1, 4)

4 = \(\frac{3}{2}\) x 1 + c ⇒ c = 4 - \(\frac{3}{2}\) = \(\frac{5}{2}.\)

∴ Required equation is y = \(\frac{3}{2}x\) + \(\frac{5}{2}\)

⇒ 3x – 2y + 5 = 0 

∴ L = 3x – 2y + 5.

No related questions found

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

...