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

Find the magnetic induction of the field at the point O of a loop with current I, whose shape is illustrated in fig. the radius a and the side b are known

Find the magnetic field at P due to the arrangement shown

If the magnitude of the resultant magnetic field inductions at the points A, B and C due to the infinitely long parallel conductors carrying currents in the direction shown are B. B and B respectively then, the ascending order of magnetic fields induction are

Consider two different ammeters in which the deflections of the needle are proportional to current. The first ammeter is connected to a resistor of resistance and the second to a resistor of unknown resistance . At first the ammeters are connected in series between points A and B [as shown in fig. a)]. In this case the readings of the ammeters are and Then the ammeters are connected parallel between A and B as shown in fig.(B) and indicated and Determine the unknown resistance of the second resistor.

A voltmeter of resistance and an ammeter of resistance are conected in series across a battery of negligible internal resistance. When a resistance R is connected in parallel to voltmeter, reading of ammeter increases three times while that of voltmeter reduces to one third. Find and in terms of R.

A galvanometer of resistance ,shunted by a resistance of 5 ohm gives a deflection of 50 divisions when joined in series with a resistance of and a 2 volt accumulator. What is the current sensitivity of the galvanometer (in div/(A) ?

A galvanometer has a current sensitivity of 1mA per division. A variable shunt is connected across the galvanometer and the combination is put in series with a resistance of and cell of internal resistance . It gives a deflection of 5 division for shunt of 5 ohm and 20 division for shunt of 25 ohm. Find the emf of cell and resistance of galvanometer.