The interior angle bisector of angle `P` for the trangle `P Q R` whose coordinates of the veetices are `P(-8,5); Q(-15 ,-19)a n dR(1,-7)` has the equation `a x+2y+c=0,` then `(a ,c)=`

Consider a triangle PQR with coordinates of its vertices as P(-8,5), Q(-15, -19), and R (1, -7). The bisector of the interior angle of P has the equation which can be written in the form ax+2y+c=0. The radius of the in circle of triangle PQR is The sum a + c is

Consider a triangle PQR with coordinates of its vertices as P(-8,5), Q(-15, -19), and R (1, -7). The bisector of the interior angle of P has the equation which can be written in the form ax+2y+c=0. The radius of the in circle of triangle PQR is

Consider a triangle PQR with coordinates of its vertices as P(-8,5), Q(-15, -19), and R (1, -7). The bisector of the interior angle of P has the equation which can be written in the form ax+2y+c=0. The distance between the orthocenter and the circumcenter of triangle PQR is

If the x-coordinate of a point P on the join of Q(22,1)a n dR(5,1,-2)i s4, then find its z- coordinate.

Find the value of x such that P Q=Q R where the coordinates of P ,\ Q and R are (6,\ -1) , (1,\ 3) and (x ,\ 8) respectively.

Construct a triangle P Q R whose perimeter is equal to 14 c m ,\ /_p=45^0\ a n d\ /_q=60^0

The x-coordinate of a point P is twice its y-coordinate. If P is equidistant from Q(2,-5)a n dR(-3,6), then find the coordinates of P.

If (x_1, y_1)&(x_2,y_2) are the ends of a diameter of a circle such that x_1&x_2 are the roots of the equation a x^2+b x+c=0 and y_1&y_2 are the roots of the equation p y^2=q y+c=0. Then the coordinates of the centre of the circle is: (b/(2a), q/(2p)) ( b/(2a), q/(2p)) (b/a , q/p) d. none of these

If p , q , r are the legths of the internal bisectors of angles A ,B , C respectively of a !ABC , then area of ABC

The product of the sines of the angles of a triangle is p and the product of their cosines is qdot Show that the tangents of the angles are the roots of the equation q x^3-p x^2+(1+q)x-p=0.