Show that the reflection of the line `px+qy+r=0` in the line `x+y+1 =0` is the line `qx+py+(p+q-r)=0,` where `p!= -q`.

If r is the geometric mean of p and q, then the line px+qy+r=0

Let P be the foot of perpendicular drawn from a point Q on the line x + y +4 = 0 to the line 5x + y = 0. If P is the image of point R(12, 0) about the line x + y = 2, then coordinates of Q are

Let P be the foot of perpendicular drawn from a point Q on the line x + y +4 = 0 to the line 5x + y = 0. If P is the image of point R(12, 0) about the line x + y = 2, then coordinates of Q are

The locus of the orthocentre of the triangle formed by the lines (1+p)x-py+p(1+p)=0 ,(1+q)x-qy+q(1+q)=0 and y=0 where p!=q is (A) a hyperbola (B) a parabola (C) an ellipse (D) a straight line

The locus of the orthocentre of the triangle formed by the lines (1+p)x-py+p(1+p)=0,(1+q)x-qy+q(1+q)=0 and y=0, where p!=*q, is (A) a hyperbola (B) a parabola (C) an ellipse (D) a straight line

Let image of the line (x-1)/(3)=(y-3)/(5)=(z-4)/(2) in the plane 2x-y+z+3=0 be L. A plane 7x+py+qz+r=0 is such that it contains the line L and perpendicular to the plane 2x-y+z+3=0 then p=1 (b) q=42p+q+r=30( d) p+q+r=0

Solve the following pair of linear equations : px + qy= p-q qx- py= p+q

Shown below are the graphs of the lines y-2x=0, x+y=6 and px+qy=r Which of these is the solution for the pair of equations x+y=6 and px+qy=r