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Find the points on the x-axis, whose distances from the x3+y4=1 are 4 units.

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Question

Find the points on the x-axis, whose distances from the `x/3 +y/4 = 1`  are 4 units.

Sum
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Solution

The given equation is: `"x"/3 + "y"/4 = 1`

multiplying by 12

4x + 3y – 12 = 0…………(i)

Suppose there is a point (a, 0) on the x-axis, then the distance of line (i) from point (a, 0) is

= `(4"a" + 0 - 12)/sqrt(16 + 9)`

= `± (4"a" - 12)/5`

∴ `± (4"a" - 12)/5 = 4`

or ± (4a – 12) = 20

By taking + ve sign 4a = 32 or a = 8

The required point on x-axis is (8, 0).

By taking – ve sign, `-(4"a" - 12)/5 = 4`

or –4a + 12 = 20

4a = –8, a = –2

The second required point is (–2, 0).

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Chapter 9: Straight Lines - EXERCISE 9.3 [Page 167]

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NCERT Mathematics [English] Class 11
Chapter 9 Straight Lines
EXERCISE 9.3 | Q 4. | Page 167

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