The distance between 1,1 and the point of intersection of two lines 2x+3y+c=0 and  3x+2y+c=0 is less than 22then

To get the point of intersection of two lines, eliminate  x32x+3y+c−23x+2y+c=06x+9y+3c−6x−4y−2c=05y+c=0y=−c5 Substitute y=−c5 in the equation 2x+3y+c=02x+3−c5+c=010x−3c+5c=010x+2c=0x=−c5 Hence, the point of intersection is −c5,−c5 Given that the distance between (1,1) and −c5,−c5 is 1+c52+1+c52<22It implies that 21+c52<22  Squaring and then simplify   21+c52<81+c52<4Apply square root on both sides −2<1+c5<2−3