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24. Find area of the triangle with vertices at the point given in each of the following:

(i) (1, 0), (6, 0), (4, 3)

(ii) (2, 7), (1, 1), (10, 8)

(iii) (–2, –3), (3, 2), (–1, –8)

0 12 Views | Posted 4 months ago
Asked by Shiksha User

  • 1 Answer

  • V

    Answered by

    Vishal Baghel | Contributor-Level 10

    4 months ago

    (i) Area of triangle is given by,

    Δ = 12 |x1y11x2y21x3y31|=12|101601431|

    = 1 2 | [ − 3 + 1 8 ] | = 1 5 2 =7.5sq. units.

    (ii) Area of the triangle is given by,

    Δ = 12 |2711111081|

    = 1 2 | [ 2 ( 1 − 8 ) − 7 ( 1 − 1 0 ) + 1 ( 8 − 1 0 ) ] |

    = 1 2 | ( 2 × ( − 7 ) − 7 × ( − 9 ) + 2 1 × ( − 2 ) ] |

    = 1 2 | [ − 1 4 + 6 3 − 2 ] | = 1 2 | 4 7 |

    = 472 =23.5 sq. units

    (iii) Area of triangle is given by,

    Δ = 12 |−2−31321−1−81|.

    = 1 2 | [ − 2 × 1 0 + 3 ( 4 ) + ( − 2 4 + 2 ) ] |

    = 1 2 | [ − 2 0 + 1 2 − 2 2 ] |

    = 1 2 | − 3 0 | = 3 0 2  = 15 sq. units.

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A = [ 1 0 0 0 α β 0 β α ] then α is (if α, β Î I)
A
alok kumar singh

|2A| = 27

8|A| = 27

Now |A| = α2–β2 = 24

α2 = 16 + β2

α2– β2 = 16

(α–β) (α+β) = 16

->α + β = 8 and

α – β = 2

->α = 5 and β = 3

If A = [ 2 1 − 1 2 ] , B = [ 1 0 1 1 ] , C = ABAT and X = AC2AT then |X| is equal to
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alok kumar singh

|A| = 3

|B| = 1

->|C| = |ABAT| = |A|B|A7| = |A|2|B|

= 9

->|X| = |A|C|2|AT|

= 3 * 92 * 3 = 9 * 92 = 729

Matrix A of order 3 × 3 is such that |A| = 2 if n = | a d j ( a d j ( a d j . . . . ( a ) ) ) | ︸ 2 0 2 4   t i m e s  then remainder when n is divided by 9 is
A
alok kumar singh

|A| = 2

a d j ( a d j ( a d j . . . . ( a ) ) ) ? 2 0 2 4   t i m e s = | A | ( n − 1 ) 2 0 2 4            

= | A | ( n − 1 ) 2 0 2 4                                             

= 2 2 2 0 2 4                                          

2 2 0 2 4 = ( 2 2 ) 2 2 0 2 2 = 4 ( 8 ) 6 7 4 = 4 ( 9 − 1 ) 6 7 4             

-> 2 2 0 2 4 ≡ 4 ( m o d   9 )

->,  2 2 0 2 4 ≡ 9 m + 4  m ¬ even

2 9 m + 4 ≡ 1 6 ⋅ ( 2 3 ) 3 m ≡ 1 6 ( m o d   9 )

7

If ∫ 0 2 ( 2 x − 2 x − x 2 ) d x = ∫ 0 1 ( 1 − 1 − y 2 − y 2 2 ) d y + ∫ 1 2 ( 2 − y 2 2 ) d y + I     t h e n     I     e q u a l s  
V
Vishal Baghel

∫ 0 2 2 x d x − ∫ 0 2 2 x − x 2 d x = ∫ 0 1 d y − ∫ 0 1 1 − y 2 d y − ∫ 0 2 y 2 2 d y + ∫ 1 2 2 d y + I

⇒ 8 3 − ∫ 0 1 1 − t 2 d t = 1 − 8 6 + 2 + I

I = 1 − ∫ 0 1 1 − t 2 d t

If  ∫ 0 2 ( 2 x − 2 x − x 2 ) d x = ∫ 0 1 ( 1 − 1 − y 2 − y 2 2 ) d y + ∫ 1 2 ( 2 − y 2 2 ) d y + I     t h e n     I     e q u a l s

A
alok kumar singh

∫ 0 2 2 x d x − ∫ 0 2 2 x − x 2 d x = ∫ 0 1 d y − ∫ 0 1 1 − y 2 d y − ∫ 0 2 y 2 2 d y + ∫ 1 2 2 d y + I

⇒ 8 3 − ∫ 0 1 1 − t 2 d t = 1 − 8 6 + 2 + I

I = 1 − ∫ 0 1 1 − t 2 d t

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