Find the distance of a point …

CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Posted by Vipul Kumar Gupta 5 years, 11 months ago
- 2 answers
Manisha Sharma 5 years, 11 months ago
Related Questions
Posted by Sahil Sahil 1 year, 4 months ago
- 2 answers
Posted by Hari Anand 6 months, 1 week ago
- 0 answers
Posted by Parinith Gowda Ms 3 months, 2 weeks ago
- 0 answers
Posted by Lakshay Kumar 1 year, 1 month ago
- 0 answers
Posted by Kanika . 1 month ago
- 1 answers
Posted by Vanshika Bhatnagar 1 year, 4 months ago
- 2 answers
Posted by Parinith Gowda Ms 3 months, 2 weeks ago
- 1 answers

myCBSEguide
Trusted by 1 Crore+ Students

Test Generator
Create papers online. It's FREE.

CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
myCBSEguide
Yogita Ingle 5 years, 11 months ago
{tex}Distance\: from\\ \:the \:point \:P(3,2)\: and\\ \:the \:origin\\=5{/tex}
Step-by-step explanation:
{tex}\boxed{Distance\: from\\ \:the \:point \:(x,y)\: and\\ \:the \:origin\: is\: \sqrt{x^{2}+y^{2}}}{/tex}
{tex}Distance\: from\\ \:the \:point \:P(3,2)=(x,y)\: and\\ \:the \:origin\\= \sqrt{3^{2}+2^{2}}\\=\sqrt{9+4}\\=\sqrt{13}\\{/tex}
Therefore,
{tex}Distance\: from\\ \:the \:point \:P(3,2)\: and\\ \:the \:origin\\=\sqrt{13}{/tex}
0Thank You