Consider a pair of perpendicular straight lines ax2+3xy−2y2−5x+5y+c=0The value of a isThe value of c isDistance between the orthocenter and the circumcenter of triangle ABC is

As the lines are perpendicular,Coefficient of x2 + Coefficient of y2 a−2=0or  a=2  As the lines are perpendicular,Coefficient of x2 + Coefficient of y2 a−2=0or  a=2Also, abc+2fgh−af2−bg2−ch2=0∴c=−3 As the lines are perpendicular,Coefficient of x2 + Coefficient of y2 a−2=0or  a=2Also, abc+2fgh−af2−bg2−ch2=0∴c=−3Hence, the given pair of lines is 2x2+3xy−2y2−5x+5y−3=0Factorizing, we get lines  x+2y−3=0 and 2x−y+1=0The point of intersection of the lines is C( 1/5 , 7/5).The points of intersection of the lines with the r-axis are A(3, 0) and 8(-1/2, 0).The orthocentre of triangle is C(1/5,7/5) and the circumcentre isthe midpoint of AB which is M(5/4, 0). ThereforeCM=(54−15)2+4925=74