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The area of a triangle with vertices (–3, 0), (3, 0) and (0, k) is 9 sq.units. The value of k will be ______. - Mathematics

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प्रश्न

The area of a triangle with vertices (–3, 0), (3, 0) and (0, k) is 9 sq.units. The value of k will be ______.

विकल्प

  • 9

  • ±3

  • – 9

  • 6

MCQ
रिक्त स्थान भरें

उत्तर

The area of a triangle with vertices (–3, 0), (3, 0) and (0, k) is 9 sq.units. The value of k will be ±3.

Explanation:

We know that, area of triangle with vertices (x1, y1), (x2, y2) and (x3, y3) is given by

Δ = `1/2|(x_1, y_1, 1),(x_2, y_2, 1),(x_3, y_3, 1)|`

∴ Area of triangle with verticles (–3, 0), (3, 0) and (0, k) is

∴ Δ = `1/2|(-3, 0, 1),(3, 0, 1),(0, "k", 1)|` = 9  ...(Given)

⇒ `[-3(-"k") - 0 + 1(3"k")]` = ±18

⇒ 6k = ±18

∴ k = `+- 18/6` = ±3

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अध्याय 4: Determinants - Exercise [पृष्ठ ८०]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
अध्याय 4 Determinants
Exercise | Q 26 | पृष्ठ ८०

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