The area of the triangle formed by the lines y=x+k,ax2+2hxy+by2=0 is
k2h2−abb+2h−a
k2h2−aba+2h−b
k2h2−aba+2h+b
k2h2−aba−2h+b
The area of the triangle formed by the lines ax2+2hxy+by2=0 and lx+my+n=0 is n2h2−abam2−2hlm+bl2Since the equation is y=x+k,l=1,m=−1,n=kTherefore, the area of the triangle is k2h2−aba(−1)2−2h(1)(−1)+b(1)2=k2h2−aba+2h+b