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प्रश्न
A point equidistant from the lines 4x + 3y + 10 = 0, 5x – 12y + 26 = 0 and 7x + 24y – 50 = 0 is ______.
पर्याय
(1, –1)
(1, 1)
(0, 0)
(0, 1)
उत्तर
A point equidistant from the lines 4x + 3y + 10 = 0, 5x – 12y + 26 = 0 and 7x + 24y – 50 = 0 is (0, 0).
Explanation:
Given equations are 4x + 3y + 10 = 0 .....(i)
5x – 12y + 26 = 0 ......(ii)
And 7x + 24y – 50 = 0 ......(iii)
Let (x1, y1) be any point equidistant from equation (i), equation (ii) and equation (iii).
Distance of (x1, y1) from equation (i)
= `|(4x_1 + 3y_1 + 10)/sqrt(16 + 9)|`
= `|(4x_1 + 3y_1 + 10)/5|`
Distance of (x1, y1) from equation (ii)
= `|(5x_1 - 12y_1 + 26)/sqrt(25 + 144)|`
= `|(5x_1 + 12y_1 + 26)/13|`
Distance of (x1, y1) from equation (iii)
= `|(7x_1 + 24y_1 - 50)/sqrt(49 + 576)|`
= `|(7x_1 + 24y_1 - 50)/25|`
If the point (x1, y1) is equidistant from the given lines, then
`|(4x_1 + 3y_1 + 10)/5| = |(5x_1 - 12y_1 + 26)/13|`
= `|(7x_1 + 2y_1 - 50)/25|`
We see that putting x1 = 0 and y1 = 0, the above relation is satisfied
i.e., `10/5 = 26/13 = 50/25` = 2
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