Question-58652 Tinku Tara June 4, 2023 Coordinate Geometry 0 Comments FacebookTweetPin Question Number 58652 by peter frank last updated on 27/Apr/19 Commented by peter frank last updated on 27/Apr/19 aandb Answered by tanmay last updated on 27/Apr/19 APis(y−0)=m(x−5)BQis(y−0)=m(x+5)solvey=m(x−5)and4x+3y=25tofindP4x+3m(x−5)=254x+3mx−15m=25x=25+15m4+3mandy=m(25+15m4+3m−5)=m(25+15m−20−15m4+3m)=(5m4+3m)soP(25+15m4+3m,5m4+3m)nowsolvey=m(x+5)and4x+3y=25tofindQ4x+3m(x+5)=254x+3mx+15m=25x=25−15m4+3m,y=m(25−15m4+3m+5)Q(25−15m4+3m,(25−15m+20+15m4+3m)×m(25−15m4+3m,45m4+3m)PQ={(25+15m4+3m)−(25−15m4+3m)}2+(45m−5m4+3m)2=900m2(4+3m)2+1600m2(4+3m)2=550m4+3m=550m−15m=20m=2035=4750m4+3m=−550m=−20−15m65m=−20m=−413 Commented by peter frank last updated on 28/Apr/19 thankyou Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Question-189720Next Next post: Question-124191 Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.