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)=` `a=11` `a=10` `c=78` `c=80`

The interior angle bisector of angle P for the trangle PQR whose coordinates of the veetices are P(-8,5);Q(-15,-19) and R(1,-7) has the equation ax+2y+c=0, then (a,c)=a=11a=10c=78c=80

The interior angle bisector of angle A for the triangle ABC whose coordinates of the vertices are A(-8,5), B(-15,-19) and C(1,-7) has the equation ax+2y+c=0 Find 'a' and '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 distance between the orthocenter and the circumcenter of triangle PQR is

In DeltaPQR , the equation of the internal angle bisector of angle Q is y = x and the equation of side PR is 3x-y=2 . If coordinates of P are (3, 2) and 2PQ = RQ , then the coordinates of Q are

Three vertices of triangle ABC are A(–1, 11), B(–9, –8) and C(15, –2). The equation of angle bisector of angle A is

If p and q are the roots of the equation x^2 - 15x + r = 0 and. p - q = 1 , then what is the value of r?

The value of p and q(p!=0,q!=0) for which p,q are the roots of the equation x^(2)+px+q=0 are (a)p=1,q=-2(b)p=-1,q=-2(c)p=-1,q=2(d)p=1,q=2

If P (x,y) and Q (1,4) are the points on the circle whose centre is :C(5,7) and radius is 5 cm, then find the locus of P.