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Question
If the slope of a line passing through the point A(3, 2) is `3/4`, then find points on the line which are 5 units away from the point A.
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Solution
Equation of the line passing through (3, 2) having slope `3/4` is given by
y – 2 = `3/4 (x - 3)`
or 4y – 3x + 1 = 0 ....(1)
Let (h, k) be the points on the line such that
(h – 3)2 + (k – 2)2 = 25 ....(2) (Why?)
Also, we have 4k – 3h + 1 = 0 ...(3) (Why?)
or k = `(3"h" - 1)/4` .....(4)
Putting the value of k in (2) and on simplifying, we get
25h2 – 150h – 175 = 0
or h2 – 6h – 7 = 0
or (h + 1)(h – 7) = 0
⇒ h = –1, h = 7
Putting these values of k in (4)
We get k = –1 and k = 5.
Therefore, the coordinates of the required points are either (–1, –1) or (7, 5).
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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) |
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(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)` |
