Let \( ABC \) be a triangle formed by the lines \( 7x - 6y + 3 = 0 \), \( x + 2y - 31 = 0 \), and \( 9x - 2y - 19 = 0 \).
Let the point \( (h, k) \) be the image of the centroid of \( \triangle ABC \) in the line \( 3x + 6y - 53 = 0 \). Then \( h^2 + k^2 + hk \) is equal to:
Step 1: The equations of the lines form a triangle, and we need to find the coordinates of the centroid of the triangle. To do this, solve the system of linear equations to find the vertices \( A \), \( B \), and \( C \) of the triangle.
Step 2: The centroid of a triangle is the average of the coordinates of its vertices. After determining the coordinates of the vertices, calculate the centroid using the formula: \[ G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \]
Step 3: Next, find the image of the centroid under the given line transformation. The image of the centroid \( G \) is the point \( (h, k) \).
Step 4: Finally, calculate \( h^2 + k^2 + hk \) using the obtained values of \( h \) and \( k \). Thus, the correct answer is (4).
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,