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
−15,5
−10,0
−10,10
−195,0
To get the point of intersection of two lines, eliminate x
32x+3y+c−23x+2y+c=06x+9y+3c−6x−4y−2c=05y+c=0y=−c5
Substitute y=−c5 in the equation 2x+3y+c=0
2x+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<22
It implies that
21+c52<22 Squaring and then simplify 21+c52<81+c52<4Apply square root on both sides
−2<1+c5<2−3<c5<1−15<c<5
Therefore, c∈(−15,5)