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प्रश्न
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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उत्तर
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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(iii) `(1, 12/5), (-3, 16/5)` |
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| Column C1 | Column C2 |
| (a) Through the point (2, 1) is | (i) 2x – y = 4 |
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(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 |
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