(S1) : If $A(5, -1)$ and $B(-2, 3)$ are two vertices of a triangle, whose orthocentre is $(0, 0)$, then its third vertex is $(-4, -7)$
and
(S2) : If positive numbers $2a, b, c$ are three consecutive terms of an A.P., then the lines $ax+by+c=0$ are concurrent at $(2, -2)$,
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 ($m_1 \cdot m_2 = -1$):
- 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)
- 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:
$ 2x - 3(5x + 13) - 13 = 0 $ $ 2x - 15x - 39 - 13 = 0 $ $ -13x - 52 = 0 $ $ -13x = 52 $ $ x = -4 $Substitute $x = -4$ back into $y = 5x + 13$:
$ y = 5(-4) + 13 = -20 + 13 = -7 $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:
$ 2b = 2a + c $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:
$ a(2) + b(-2) + c = 0 $ $ 2a - 2b + c = 0 $Now, let's use the A.P. condition ($2b = 2a + c$) to verify this.
Rearrange the A.P. condition to solve for $c$:
$ c = 2b - 2a $Substitute this expression for $c$ into the equation $2a - 2b + c = 0$:
$ 2a - 2b + (2b - 2a) = 0 $ $ 0 = 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.
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Important Questions from Coordinate Geometry
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