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तामिळनाडू बोर्ड ऑफ सेकेंडरी एज्युकेशनएस.एस.एल.सी. (इंग्रजी माध्यम) इयत्ता १०

A person standing at a junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 seek to reach the path whose equation is 6x – 7y + 8 = 0 in the least - Mathematics

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प्रश्न

A person standing at a junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 seek to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find the equation of the path that he should follow.

बेरीज

उत्तर

Two straight paths will intersect at one point.

Solving this equation

2x – 3y + 4 = 0

                    2x – 3y = – 4     ...(1)
                    3x + 4y = 5       ...(2)
(1) × 4 ⇒    8x – 12y = – 16  ...(3)
(2) × 3 ⇒    9x + 12y = 15    ...(4)
(3) + (4) ⇒  17x         = – 1
                       x         = `(-1)/17`

Substitute the value of x = `(-1)/17` in (2)

`3(- 1/17) + 4y` = 5

⇒ `-3/17 + 4y` = 5

4y = `5 + 3/17`

= `(85 + 3)/17`

4y = `88/17`

⇒ y = `88/(17 xx 4)`

= `22/17`

The point of intersection is `(-1/17, 22/17)`

Any equation perpendicular to 6x – 7y + 8 = 0 is 7x + 6y + k = 0

It passes through `(-1/17, 22/17)`

`7(-1/17) + 6(22/17) + "k"` = 0

Multiply by 17

– 7 + 6(22) + 17k = 0

– 7 + 132 + 17k = 0

17k = – 125

⇒ k = `- 125/17`

The equation of a line is `7x + 6y - 125/17` = 0

119x + 102y – 125 = 0

∴ Equation of the path is 119x + 102y – 125 = 0

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General Form of a Straight Line
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पाठ 5: Coordinate Geometry - Unit Exercise – 5 [पृष्ठ २३७]

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सामाचीर कलवी Mathematics [English] Class 10 SSLC TN Board
पाठ 5 Coordinate Geometry
Unit Exercise – 5 | Q 10 | पृष्ठ २३७
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