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).

The sides of a rhombus ABCD are parallel to the lines y = x + 2; y = 7x + 3. If the diagonals of the rhombus intersect at the point (1, 2) and the vertex A is on y – axis then A =

The quadrilateral formed by the lines + y = 0; x + y = 0; x + y = 1; x + y + 1 = 0 is

If 7x – y + 3 = 0; x + y – 3 = 0 are tow sides of an isosceles triangle and the third side passes through (1, 0) then the equation of the third side is

So here we used the concept of the triangles, and the equations of the lines to solve this problem. We know that the equation of a straight line in slope-intercept form is given as y = mx+c, so we used that here. So the equation of the third side is x + 3y - 1 = 0.

All points lying inside the triangle formed by the points (1, 3) (5, 0) and (-1, 2) satisfy

In Young's double slit experiment with a monochromatic light of wavelength , the fringe width is found to be 0.4 mm. When the slits are now illuminated with a light of wavelength the fringe width will be

The magnetic field due to a current carrying square loop of side a at a point located symmetrically at a distance of a/2 from its centre (as shown is)

A long wire carrying current i is bent to form a right angle as shown in figure .Then magnetic induction at point P on the bisector of this angle at a distance r from the vertex is