Consider a pair of perpendicular straight lines `ax^(2)+3xy-2y^(2)-5x+5y+c=0`.
then the value of c is

Options

A. -3
B. `9//2`
C. `8//3`
D. `7//4`

Answer

4

Explanation

As the lines are perpendicular, Coefficient of `x^(2)+` Coefficient of `y^(2)=0`
`:.a-2=0`
or `a=2`
Also , `abc+2fgh-af^(2)-bg^(2)-ch^(2)=0`
`:. c=-3`
Hence , the given pair of lines is
`2x^(2)+3xy-2y^(2)-5x+5y-3=0`
Factorizing , we get lines
`x+2y-3=0and 2x-y+1=0`

The point of intersection of the lines is C `(1//5,7//5)`.
The points of intersection of the lines with the x- axis are A(3,0) and B `(-1//2,0)`.
The orthocenter of triangle is C `(1//5,7//5)` and the circumcenter is the midpoint of AB which is M `(5//4,0)`. Therefore,
CM`sqrt(((5)/(4)-(1)/(5))^(2)+(49)/(25))=(7)/(4)`