Given:
- In (S1), A(5,−1) and B(−2,3) are two vertices of a triangle and the orthocentre is H(0,0).
- In (S2), 2a,b,c are three consecutive terms of an A.P. and the family of lines is ax+by+c=0.
Find: Whether statements (S1) and (S2) are correct.
For (S1), verify whether the third vertex (−4,−7) gives orthocentre at the origin.
Slope of AB is
−2−53−(−1)=−74=−74
So the altitude from C must have slope
47
Now slope of the line joining C(−4,−7) to H(0,0) is
0−(−4)0−(−7)=47
Hence, CH is perpendicular to AB.
Next, slope of BC is
−4−(−2)−7−3=−2−10=5
So the altitude from A must have slope
−51
The slope of AH is
0−50−(−1)=−51
Hence, AH is perpendicular to BC.
Also, slope of CA is
5−(−4)−1−(−7)=96=32
So the altitude from B must have slope
−23
The slope of BH is
0−(−2)0−3=−23
Hence, BH is perpendicular to CA.
Therefore, all three altitudes pass through H(0,0), so (S1) is correct.
For (S2), since 2a,b,c are consecutive terms of an A.P.,
b−2a=c−b
Therefore,
2b=2a+c
Now test the point (2,−2) in the line
ax+by+c=0
Substituting x=2 and y=−2,
2a−2b+c=0
Using 2b=2a+c,
2a−(2a+c)+c=0
So the equation is satisfied identically. Hence every such line passes through (2,−2), so (S2) is correct.
Therefore, both statements are correct. The correct option is A.