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Question
The reflection of the point (4, −13) about the line 5x + y + 6 = 0 is
Options
(−1, −14)
(3, 4)
(0, 0)
(1, 2)
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Solution
Let the reflection point be A(h, k)
Now, the mid point of line joining (h, k) and (4, −13) will lie on the line 5x + y + 6 = 0
\[\therefore 5\left( \frac{h + 4}{2} \right) + \frac{k - 13}{2} + 6 = 0\]
\[ \Rightarrow 5h + 20 + k - 13 + 12 = 0\]
\[ \Rightarrow 5h + k + 19 = 0 . . . . . \left( 1 \right)\]
Now, the slope of the line joining points (h, k) and (4,−13) are perpendicular to the line 5x + y + 6 = 0.
slope of the line = −5
slope of line joining by points (h, k) and (4,−13)
\[\frac{k + 13}{h - 4}\]
\[\therefore \frac{k + 13}{h - 4}\left( - 5 \right) = - 1\]
\[ \Rightarrow 5k - h + 69 = 0 . . . . . \left( 2 \right)\]
Solving (1) and (2), we get
h = −1 and k = −14
Hence, the correct answer is option (a).
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| Column C1 | Column C2 |
| (a) The coordinates of the points P and Q on the line x + 5y = 13 which are at a distance of 2 units from the line 12x – 5y + 26 = 0 are |
(i) (3, 1), (–7, 11) |
| (b) The coordinates of the point on the line x + y = 4, which are at a unit distance from the line 4x + 3y – 10 = 0 are |
(ii) `(- 1/3, 11/3), (4/3, 7/3)` |
| (c) The coordinates of the point on the line joining A (–2, 5) and B (3, 1) such that AP = PQ = QB are |
(iii) `(1, 12/5), (-3, 16/5)` |
