Given two lines:
Equation 1: \( x + 2y - 31 = 0 \)
Equation 2: \( 9x - 2y - 19 = 0 \)
Solving these two equations, we find the points of intersection:
Intersection points are: \( (9,11) \), \( (3,4) \), \( (5,13) \)
The centroid of \( \triangle ABC \) is calculated as:
\[ \text{Centroid} = \left( \frac{17}{3}, \frac{28}{3} \right) \]
Since the image of \( \triangle ABC \) is reflected about the line:
\[ 2x + 6y - 53 = 0 \]
The centroid of the reflected triangle will also be the reflection of the original centroid across this line.
Using the formula for reflection of a point \( (x, y) \) about the line \( ax + by + c = 0 \):
\[ x' = x - \frac{2a(ax + by + c)}{a^2 + b^2}, \quad y' = y - \frac{2b(ax + by + c)}{a^2 + b^2} \]
Substituting \( (x, y) = \left( \frac{17}{3}, \frac{28}{3} \right) \) and \( a = 2 \), \( b = 6 \), \( c = -53 \):
\[ \frac{x - \frac{17}{3}}{2} = \frac{-2\left(2\left(\frac{17}{3}\right) + 6\left(\frac{28}{3}\right) - 53\right)}{2^2 + 6^2} \]
Solving for \( h \) and \( k \):
\[ h = 3, \quad k = 4 \]
Finally, computing:
\[ h^2 + k^2 + hk = (h + k)^2 - hk \]
\[ = 49 - 12 = 37 \]
Let a line passing through the point $ (4,1,0) $ intersect the line $ L_1: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} $ at the point $ A(\alpha, \beta, \gamma) $ and the line $ L_2: x - 6 = y = -z + 4 $ at the point $ B(a, b, c) $. Then $ \begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix} \text{ is equal to} $
20 mL of sodium iodide solution gave 4.74 g silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide solution is _____ M. (Nearest Integer value) (Given : Na = 23, I = 127, Ag = 108, N = 14, O = 16 g mol$^{-1}$)