Statement S1 Analysis: Triangle Orthocentre and Vertices

We are given two vertices of a triangle, $A(5, -1)$ and $B(-2, 3)$, and its orthocentre $H(0, 0)$. We need to find the third vertex $C(x, y)$ and verify if it is $(-4, -7)$.

The orthocentre is the intersection point of the altitudes of a triangle. An altitude from a vertex is perpendicular to the opposite side.

• The line segment $AH$ connects vertex A to the orthocentre H. The altitude from vertex B is perpendicular to side $AC$. Thus, $BH \perp AC$.
• Similarly, the altitude from vertex A is perpendicular to side $BC$. Thus, $AH \perp BC$.

First, let's find the slopes:

• Slope of $AH$, $m_{AH} = \frac{-1 - 0}{5 - 0} = -\frac{1}{5}$.
• Slope of $BH$, $m_{BH} = \frac{3 - 0}{-2 - 0} = -\frac{3}{2}$.
• Slope of $BC$, $m_{BC} = \frac{y - 3}{x - (-2)} = \frac{y - 3}{x + 2}$.
• Slope of $AC$, $m_{AC} = \frac{y - (-1)}{x - 5} = \frac{y + 1}{x - 5}$.

Using the perpendicularity conditions ($m_1 \cdot m_2 = -1$):

1. Condition $AH \perp BC$: $m_{AH} \cdot m_{BC} = -1$ $ (-\frac{1}{5}) \left( \frac{y - 3}{x + 2} \right) = -1 $ $ \frac{y - 3}{x + 2} = 5 $ $ y - 3 = 5(x + 2) $ $ y - 3 = 5x + 10 $ $ 5x - y + 13 = 0 $ (Equation 1)
2. Condition $BH \perp AC$: $m_{BH} \cdot m_{AC} = -1$ $ (-\frac{3}{2}) \left( \frac{y + 1}{x - 5} \right) = -1 $ $ \frac{y + 1}{x - 5} = \frac{2}{3} $ $ 3(y + 1) = 2(x - 5) $ $ 3y + 3 = 2x - 10 $ $ 2x - 3y - 13 = 0 $ (Equation 2)

Now, we solve the system of linear equations (Equation 1 and Equation 2):

From Equation 1, $ y = 5x + 13 $. Substitute this into Equation 2:

Substitute $x = -4$ back into $y = 5x + 13$:

The third vertex $C$ is $(-4, -7)$. Therefore, statement S1 is correct.

Statement S2 Analysis: Arithmetic Progression and Line Concurrency

We are given that $2a, b, c$ are three consecutive terms of an Arithmetic Progression (A.P.). This implies the middle term is the average of the other two, or $2 \times (\text{middle term}) = \text{sum of first and third}$.

Mathematically, this means:

We need to check if the line $ax + by + c = 0$ is concurrent at the point $(2, -2)$. For concurrency, the point must satisfy the line equation.

Substitute $(x, y) = (2, -2)$ into the line equation:

Now, let's use the A.P. condition ($2b = 2a + c$) to verify this.

Rearrange the A.P. condition to solve for $c$:

Substitute this expression for $c$ into the equation $2a - 2b + c = 0$:

Since the equation $0 = 0$ is always true, the point $(2, -2)$ lies on the line $ax + by + c = 0$ for any positive numbers $2a, b, c$ that form an A.P. Thus, the lines are concurrent at $(2, -2)$.

Therefore, statement S2 is correct.

Conclusion

Both statement S1 and statement S2 have been verified as correct.