Given the family of lines `a (3x +4y+6)+b(x + y + 2) = 0` The line of the family situated at the greatest distance from the point P(2,3) has equation

Given a family of lines a(2x +y+4) + b(x-2y-3)=0 .The number of lines belonging to the family at a distance of sqrt(10) from point (2,-3) is

The distance of the point (-2,6) from the line 3x - 4y - 10=0

Consider the line L:(x-1)/(2)=(y-1)/(-3)=(z+10)/(8) and a family of planes P containing the line L. The member of the family of planes P which is situated at a maximum distance from A(1, 0, 0) will be

A family of lines is given by (1+2lambda)x+(1-lambda)y+lambda=0 being a parameter . The line belonging to this family at the maximum distance from the point (1,4) is ax+by+c=0 , then find the value of (a+b+c)/16

Let L be the line belonging to the family of straight lines (a+2b) x+(a-3b)y+a-8b =0, a, b in R, which is the farthest from the point (2, 2). If L is concurrent with the lines x-2y+1=0 and 3x-4y+ lambda=0 , then the value of lambda is

Consider a family of lines (4a+3)x-(a+1)y-(2a+1)=0 where a in R a member of this family with positive gradient making an angle of 45 with the line 3x-4y=2 , is

If 2a^2 - 7ab -ac +3b^2-2bc-c^2=0 then the family of lines ax + by + c=0 are either concurrent at the point P(x_1,y_1) or at the point Q(x_2, y_2) . Find the distance of the origin from the line passing through the points P and Q, the distance being measured parallel to the line 3x-4y=2 .

The shortest distance from the line 3x+4y=25 to the circle x^2+y^2=6x-8y is equal to :

Find the equations of the lines through the point of intersection of the lines x-y+1=0a n d2x-3y+5=0 whose distance from the point (3,2) is 7/5.

Find the equations of the lines through the point of intersection of the lines x -y+1=0 and 2x-3y+5= 0 whose distance from the point (3, 2) is 7/5