We are given two lines representing equal sides of an isosceles triangle. The goal is to find the sum of all possible distinct values of the slope \( m \) of the third side.
To solve for the third side, we first calculate the intersection points of the given lines:
1. The first line is \( -x + 2y = 4 \), which can be rewritten as: \[ y = \frac{x + 4}{2} \]
2. The second line is \( x + y = 4 \), which simplifies to: \[ y = 4 - x \]
Now, we find the intersection of these two lines by solving the system of equations:
\[ \frac{x + 4}{2} = 4 - x \]
Solve this equation to find the point of intersection. Then, calculate the slopes of the lines formed by the points of intersection with the third side. Finally, sum all distinct possible slopes of the third side.
Answer: The sum of all possible distinct values of \( m \) is \( \boxed{6} \).
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} $
Let \( T_r \) be the \( r^{\text{th}} \) term of an A.P. If for some \( m \), \( T_m = \dfrac{1}{25} \), \( T_{25} = \dfrac{1}{20} \), and \( \displaystyle\sum_{r=1}^{25} T_r = 13 \), then \( 5m \displaystyle\sum_{r=m}^{2m} T_r \) is equal to:
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): The density of the copper ($^{64}Cu$) nucleus is greater than that of the carbon ($^{12}C$) nucleus.
Reason (R): The nucleus of mass number A has a radius proportional to $A^{1/3}$.
In the light of the above statements, choose the most appropriate answer from the options given below: