Home
Class 12
MATHS
From a point P=(3, 4) perpendiculars PQ ...

From a point` P=(3, 4)` perpendiculars `PQ` and `PR` are drawn to line `3x +4y -7=0` and a variable line `y -1= m (x-7)` respectively then maximum area of triangle `PQR` is :

Promotional Banner

Similar Questions

Explore conceptually related problems

From a point P perpendicular tangents PQ and PR are drawn to ellipse x^(2)+4y^(2) =4 , then locus of circumcentre of triangle PQR is

From a point P perpendicular tangents PQ and PR are drawn to ellipse x^(2)+4y^(2) =4 , then locus of circumcentre of triangle PQR is

From a point P perpendicular tangents PQ and PR are drawn to ellipse x^(2)+4y^(2) =4 , then locus of circumcentre of triangle PQR is

The pedal points of a perpendicular drawn from origin on the line 3x+4y-5=0, is

From a point P (lamda, lamda, lamda) perpendiculars PQ and PR are drawn respectively on the lines y =x, z = 1 and y = -x, z = -1 . If P is such that angleQPR is a right angle, then the possible value (s) of lamda is (are)

lf from point P(4,4) perpendiculars to the straight lines 3x+4y+5=0 and y=mx+7 meet at Q and R area of triangle PQR is maximum, then m is equal to

From a point P(lambda,lambda,lambda) , perpendicular PQ and PR are drawn respectively on the lines y=x, z= 1 and y=-x, z=-1 .If P is such that angleQPR is a right angle, then the possible value(s) of lambda is/(are)

If from point P(4,4) perpendiculars to the straight lines 3x+4y+5=0 and y=mx+7 meet at Q and R area of triangle PQR is maximum,then m is equal to

Find the length of perpendicular drawn from point (2,-1) to the line 3x+4y-11=0 .

Find the length of perpendicular drawn from point (2,-1) to the line 3x+4y-11=0 .