find the point on the straight line 3x+y+4=0 which is equidistant from the points (-5,6) and (3,2) Share with your friends Share 0 Neha Sethi answered this Dear student 3x+y+4=0 -5,6 3,2Let the point be P(x,y)Then 3x+y+4=0 ....(1)Since the line (1) is equidistant from (-5,6) and (3,2)So, (x+5)2+(y-6)2=(x-3)2+(y-2)2⇒x2+25+25x+y2+36-12y=x2+9-6x+y2+4-4y⇒31x-8y+48=0 ...(2)Solving (1) and (2), we getx=-1611 and y=411So, P=-1611,411 Regards 1 View Full Answer