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महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता १२ वी

Find the joint equation of the line passing through the origin and perpendicular to the lines x + 2y = 19 and 3x + y = 18

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प्रश्न

Find the joint equation of the line passing through (-1, 2) and perpendicular to the lines  x + 2y + 3 = 0 and 3x - 4y - 5 = 0

बेरीज
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उत्तर

Let L1 and L2 be the lines passing through the origin and perpendicular to the lines x + 2y + 3 = 0 and 3x - 4y - 5 = 0 respectively.

Slopes of the lines x + 2y + 3 = 0 and 3x - 4y - 5 = 0 are `-1/2` and `- 3/-4 = 3/4`   respectively.

∴ slopes of the lines L1 and L2 are `2` and `(-4)/3` respectively.

Since the lines L1 and L2 pass through the point (-1, 2), their equations are

∴ (y - y1) = m(x - x1)

∴ (y - 2)= 2(x + 1)

⇒ y - 2 = 2x + 2

2x - y + 4 = 0 and 

∴ (y - 2) = `((-4)/3)`(x + 1)

⇒ 3y - 6 = (- 4)(x + 1)

⇒ 3y - 6 = - 4x - 4

⇒ 4x + 3y - 6 + 4 = 0

4x + 3y - 2 = 0

their combined equation is

∴ (2x - y + 4)(4x + 3y - 2) = 0

∴ 8x2 + 6xy - 4x - 4xy - 3y2 + 2y + 16x + 12y - 8 = 0

∴ 8x2 + 2xy + 12x - 3y2 + 14y - 8 = 0

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पाठ 4: Pair of Straight Lines - Miscellaneous Exercise 4 [पृष्ठ १३१]

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बालभारती Mathematics and Statistics 1 (Arts and Science) [English] Standard 12 Maharashtra State Board
पाठ 4 Pair of Straight Lines
Miscellaneous Exercise 4 | Q 1.07 | पृष्ठ १३१

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