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A point equidistant from the lines 4x + 3y + 10 = 0, 5x – 12y + 26 = 0 and 7x + 24y – 50 = 0 is
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- (1, –1)
- (1, 1)
- (0, 0)
- (0, 1)
Correct Option: C
Here, it is clear that distance of the given lines from (0, 0) is equal.
| d1 = |  | | |
| √1 + m² |
| = | | | |
| 5 |
= 2 units
| d2 = | | | |
| √5² + 12² |
| = | | | |
| 13 |
= 2 units
| d3 = | | | |
| √7² + 24² |
| = | | | |
| √625 |
| = | | | |
| √25 |
d3 = 2
