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प्रश्न
The reflection of the point (4, – 13) about the line 5x + y + 6 = 0 is ______.
विकल्प
(– 1, – 14)
(3, 4)
(0, 0)
(1, 2)
उत्तर
The reflection of the point (4, – 13) about the line 5x + y + 6 = 0 is (– 1, – 14).
Explanation:
Let (h, k) be the point of reflection of the given point (4, – 13) about the line 5x + y + 6 = 0.
The mid-point of the line segment joining points (h, k) and (4, – 13) is given by
`(h + 4)/2, (k - 13)/2` (Why?)
This point lies on the given line, so we have
`5 (h + 4)/2 + (k - 13)/2 + 6` = 0
or 5 h + k + 19 = 0 .....(1)
Again the slope of the line joining points (h, k) and (4, –13) is given by `(k + 13)/(h - 4)`.
This line is perpendicular to the given line
Hence `(-5) (k + 3)/(h - 4)` = –1 (why?)
This gives 5k + 65 = h – 4
or h – 5k – 69 = 0 ....(2)
On solving (1) and (2)
We get h = –1 and k = –14.
Thus the point (–1, – 14) is the reflection of the given point.
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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)` |
The equation of the line through the intersection of the lines 2x – 3y = 0 and 4x – 5y = 2 and
| Column C1 | Column C2 |
| (a) Through the point (2, 1) is | (i) 2x – y = 4 |
| (b) Perpendicular to the line (ii) x + y – 5 = 0 x + 2y + 1 = 0 is |
(ii) x + y – 5 = 0 |
| (c) Parallel to the line (iii) x – y –1 = 0 3x – 4y + 5 = 0 is |
(iii) x – y –1 = 0 |
| (d) Equally inclined to the axes is | (iv) 3x – 4y – 1 = 0 |
