find the point on the straight line 3x+y+4=0 which is equidistant from the points (-5,6) and (3,2)

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)2x2+25+25x+y2+36-12y=x2+9-6x+y2+4-4y31x-8y+48=0   ...(2)Solving (1) and (2), we getx=-1611 and y=411So, P=-1611,411
Regards

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