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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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Question

A point equidistant from the lines 4x + 3y + 10 = 0, 5x – 12y + 26 = 0 and 7x + 24y – 50 = 0 is ______.

Options

  • (1, –1)

  • (1, 1)

  • (0, 0)

  • (0, 1)

MCQ
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Solution

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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Chapter 10: Straight Lines - Exercise [Page 182]

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NCERT Exemplar Mathematics Exemplar [English] Class 11
Chapter 10 Straight Lines
Exercise | Q 38 | Page 182

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