Question ID: #756
Among the statements:
(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 $(2, -2)$,
- (1) Only (S1) is correct
- (2) Only (S2) is correct
- (3) Both are incorrect
- (4) Both are correct
Solution:
For Statement (S1):
Let the third vertex be $C(h, k)$.
Since $O(0,0)$ is the orthocentre, $AO \perp BC$ and $BO \perp AC$.
Slope of $AO$, $m_{AO} = \frac{0 – (-1)}{0 – 5} = -\frac{1}{5}$.
Slope of $BC$, $m_{BC} = \frac{k – 3}{h – (-2)} = \frac{k – 3}{h + 2}$.
Since $AO \perp BC$, $m_{AO} \cdot m_{BC} = -1$:
$$ \left(-\frac{1}{5}\right) \left(\frac{k – 3}{h + 2}\right) = -1 \Rightarrow k – 3 = 5(h + 2) \Rightarrow 5h – k + 13 = 0 \quad \dots(1) $$
Slope of $BO$, $m_{BO} = \frac{0 – 3}{0 – (-2)} = -\frac{3}{2}$.
Slope of $AC$, $m_{AC} = \frac{k – (-1)}{h – 5} = \frac{k + 1}{h – 5}$.
Since $BO \perp AC$, $m_{BO} \cdot m_{AC} = -1$:
$$ \left(-\frac{3}{2}\right) \left(\frac{k + 1}{h – 5}\right) = -1 \Rightarrow 3(k + 1) = 2(h – 5) \Rightarrow 3k + 3 = 2h – 10 \Rightarrow 2h – 3k – 13 = 0 \quad \dots(2) $$
Multiplying eq(2) by 5 and eq(1) by 2 to solve for $h$ and $k$:
$10h – 2k + 26 = 0$
$10h – 15k – 65 = 0$
Subtracting gives $13k + 91 = 0 \Rightarrow k = -7$.
Substitute $k = -7$ in (1):
$5h – (-7) + 13 = 0 \Rightarrow 5h + 20 = 0 \Rightarrow h = -4$.
Vertex $C$ is $(-4, -7)$. Statement 1 is Correct.
For Statement (S2):
Given $2a, b, c$ are in A.P.
$$ 2b = 2a + c \Rightarrow c = 2b – 2a $$
The equation of the line is $ax + by + c = 0$.
Substitute $c$:
$$ ax + by + (2b – 2a) = 0 $$
$$ a(x – 2) + b(y + 2) = 0 $$
For this to hold for all $a, b$, we must have $x – 2 = 0$ and $y + 2 = 0$.
$$ x = 2, y = -2 $$
The lines are concurrent at $(2, -2)$. Statement 2 is Correct.
Ans. (4)
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