The alternate method to solve this will be using u by method. It is method used in differentiation when we have a condition of numerator and denominator.

Locus of the foot of perpendicular fr om (-2,3) to a variable line with intercept 2 is

Vertices of a triangle are (4, 3) , Then the locus of the orthocenter is

A line through A(-5, -4) meets the line x+3y+2=0, 2x+y+4=0 and x-y-5=0 at and respectively. If then the equation of the line is

If as well as are in G.P with the same common ratio, then the points .

The range of for which the points and lie on opposite side of the line 2x+3y=6 is

The internal bisectors of the a , having the sides BC=3, CA=5 and AB =4 meet the sides BC, CA and AB in D, E and F respectively, the area of is

A straight line with negative slope passes through the point (8, 2) and cuts the positive coordinate axes at points P and Q As varies, the absolute minimum value of OP+OQ= where is the origin

Vertices of a variable triangle are (3, 4) and where Locus of its orthocenter is

If (0,0), (a, 2), (2, b) form the vertices of an equilateral triangle, where a and b not lie between 0 and 2, then the value of 4(ab)‐ab equals

The lines and are concurrent if

Equation of the line perpendicular to 4x + 7y + 9 = 0 and such that the triangle formed by it with the coordinates axes forms an area of 3.5 sq. units is

(0, 0) is the foot of the perpendicular from (4, 2) on a straight line. The equation of the line is

The orthocentre of the triangle having vertices at (2, 3) (2, 5) (4, 3) is

Each side of square is length 5. The centre of square is (3, 7) and one of the diagonals is parallel to y = x. Then the coordinates of its vertices are

Therefore, the coordinates of vertices of square are (1, 5) (1, 9) (5, 9) (5, 5).