If a point `(x,y)` is equidistant from the `Q(9,8)` and `S(17,8)`,then (i) x+y=13 (ii) x-13=0 (iii) y-13=0 (iv)x-y=13

If a point (x,y) is equidistant from the points A(9,8) and B(17,8), then the relation between x and y is :

Show that the points (1, 1, 1) and (-3, 0, 1) are equidistant from the plane 3x+4y-12 z+13=0.

The locus of a point "P" ,if the join of the points (2,3) and (-1,5) subtends right angle at "P" is x^(2)+y^(2)-x-8y+13=0 x^(2)-y^(2)-x+8y+3=0 x^(2)+y^(2)-4x-4y=0,(x,y)!=(0,4)&(4,0) x^(2)+y^(2)-x-8y+13=0,(x,y)!=(2,3)&(-1,5)

The lines x+2y-5=0, 2x-3y+4=0,6x+4y-13=0

Find the value of (x+y) and (x-y) , if (i) 15x-17y=11 , 17x-15y=21 . (ii) 12x+13y=29 , 13x+12y=21 .

Show that the points (1,-1,3) and (3, 4, 3) are equidistant from the plane 5x +2y + -7z + 8 = 0.

Show that the points (-3,0,1) and (1,1,1) are equidistant from the plane 3x+4y-12z+13=0 .

The equation of the plane through the point (1,2,-3) which is parallel to the plane 3x-5y+2z=11 is given by (A) 3x-5y+2z-13=0 (B) 5x-3y+2z+13=0 (C) 3x-2y+5z+13=0 (D) 3x-5y+2z+13=0

The perpendicular distance from (1,2) to the straight line 12x+5y=7 is (i)(15)/(13) (ii) (12)/(13) (iii) (5)/(13) (iv) (7)/(13)