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
839 views
in Straight Lines by (26.9k points)
closed by

Find the distance of (2, 3) from the line x + 4y = 5

1 Answer

+1 vote
by (30.0k points)
selected by
 
Best answer

Given line is x + 4y – 5 = 0

Here, A = 1, B = 4, C = – 5 and the given point is (x1, y1) = (2, 3)

Distance, d = \(\bigg|\frac{A_1 x + B_1 y + C_1}{\sqrt{A^2 + B^2}}\bigg|\)

= \(\bigg|\frac{1.2 + 4.3 + 5}{\sqrt{1^2 + 4^2}}\bigg|\)

= \(\frac{9}{\sqrt{17}}\)

= \(\frac{9\sqrt{17}}{17}\) units

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

...