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A Point Equidistant from the Line 4x + 3y + 10 = 0, 5x − 12y + 26 = 0 and 7x+ 24y − 50 = 0 is - Mathematics

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प्रश्न

A point equidistant from the line 4x + 3y + 10 = 0, 5x − 12y + 26 = 0 and 7x+ 24y − 50 = 0 is

विकल्प

  • (1, −1)

  •  (1, 1)

  • (0, 0)

  • (0, 1)

MCQ

उत्तर

Let the coordiantes of the point be (a, b)
Now, the distance of the point (a, b) from 4x + 3y + 10 = 0 is given by

\[\left| \frac{4a + 3b + 10}{\sqrt{4^2 + 3^2}} \right|\]

\[ = \left| \frac{4a + 3b + 10}{5} \right|\]

Again, the distance of the point (a, b) from 5x − 12y + 26 = 0 is given by

\[\left| \frac{5a - 12b + 26}{\sqrt{5^2 + \left( - 12 \right)^2}} \right|\]

\[ = \left| \frac{5a - 12b + 26}{13} \right|\]

Again, the distance of the point (a, b) from 7x + 24y − 50 = 0 is is given by

\[\left| \frac{7a + 24b - 50}{\sqrt{7^2 + \left( 24 \right)^2}} \right|\]

\[ = \left| \frac{7a + 24b - 50}{25} \right|\]

Now,

\[\left| \frac{4a + 3b + 10}{5} \right| = \left| \frac{5a - 12b + 26}{13} \right| = \left| \frac{7a + 24b - 50}{25} \right|\]

Only a = 0 and b = 0 is satisfying the above equation
Hence, the correct answer is option (c).

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अध्याय 23: The straight lines - Exercise 23.21 [पृष्ठ १३५]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 23 The straight lines
Exercise 23.21 | Q 32 | पृष्ठ १३५

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