Question

please solve these mcqs
no useless answers please........

all the best......❤️❤️​

Answer 1

AnsWer :

• a = 1 / 2
• b = - 1 / 2

• k = 1
• k = 5

QuestioN :

If the point A( - 2 , 1 ) , B( a , b ) and C( 4 , 1 ) are collinear and a - b = 1. Find the value of a , b.

To FinD :

The value of a , b.

SolutioN :

Let,

• A( - 2 , 1 )
• B( a , b )
• C( 4 , - 1 )

For 'X'

• x1 = - 2
• x2 = a
• x3 = 4

For 'Y'

• y1 = 1
• y2 = b
• y3 = - 1

Now, we have Formula

$\tt \dagger \: \: \: \: \:Area \: of \: \triangle \: ABC- ( Given )$

$\tt \dagger \: \: \: \: \: \dfrac{1}{2} \Bigg|x_1(y_2 -y_1 )+x_2(y_3-y_1) +x_3(y_1-y_2)\Bigg|$

Now, Putting given values.

$\tt \implies0 = \dfrac{1}{2} \Bigg|( - 2)(b + 1)+(a)( - 1 - 1) +(4)(1-b)\Bigg|$

$\tt : \implies0 = \Bigg|( - 2b - 2)+( -2a) +(4-4b)\Bigg|$

$\tt : \implies0 = \Bigg|- 2b - 2 -2a+4-4b\Bigg|$

$\tt : \implies0 = \Bigg|2 -2a-6b\Bigg|$

$\tt : \implies0 = 2 + 2a + 6b$

$\tt : \implies0 = 1 + a + 3b$

$\tt : \implies a + 3b = - 1 - (1)$

$\tt : \implies a - b = 1 - (2)$

Form, subtraction equation ( 2 ) and ( 1 )

$\tt : \implies \cancel4b = - \cancel 2.$

$\tt : \implies b = - \dfrac{1}{2}$

Now, Putting the value of b in equation ( 2 )

$\tt : \implies a - b = 1 .$

$\tt : \implies a + \dfrac{1}{2} = 1 .$

$\tt : \implies \dfrac{2a + 1}{2} = 1 .$

$\tt : \implies 2a + 1 = 2$

$\tt : \implies 2a = 2 - 1$

$\tt : \implies 2a = 1.$

$\tt : \implies a = \dfrac{1}{2}$

Therefore, the value of a is 1 / 2 and b is - 1 / 2.

______________________________

VerificatioN :

Taking Equation ( 1 )

$\tt \dagger \: \: \: \: \: a + 3b = - 1.$

• a = 1 / 2.
• b = - 1 / 2.

$\tt : \implies a + 3b = - 1.$

$\tt : \implies \dfrac{1}{2} + 3 \times - \dfrac{1}{2} = - 1$

$\tt : \implies \dfrac{1}{2} - \dfrac{3}{2} = - 1$

$\tt : \implies \dfrac{1 - 3}{2} = - 1$

$\tt : \implies \dfrac{ - 2}{2} = - 1$

$\tt : \implies - 1= - 1$

______________________________

Taking Equation ( 2 )

$\tt \dagger \: \: \: \: \: a - b = 1 - (2)$

• a = 1 / 2
• b = - 1 / 2.

$\tt : \implies a - b = 1$

$\tt : \implies \dfrac{1}{2} - - \dfrac{1}{2} = 1$

$\tt : \implies \dfrac{1 + 1}{2} = 1$

$\tt : \implies \dfrac{2}{2} = 1$

$\tt : \implies1 = 1.$

Hence Verified.

______________________________

Q8.

If the point P( k - 1 , 2 ) is equidistant from point A( 3 , k ) and B( k , 5 ) , Find the value of k.

To FinD :

Thev value of k.

SolutioN :

Let,

• A( 3 , k )
• B( k , 5 )
• P( k - 1 , 2 )

P is equidistant from point A and B.

so, By ( Distance Formula )

→ AP = BP

$\tt : \implies \sqrt{{\Big(k - 1 - 3\Big)}^2+{ \Big(2 - k\Big)^2}}= \sqrt{{\Big(k - 1 - k\Big)}^2+{ \Big(2 - 5\Big)^2}}$

$\tt : \implies {\Big(k - 1 - 3\Big)}^2+{ \Big(2 - k\Big)^2}= {\Big(k - 1 - k\Big)}^2+{ \Big(2 - 5\Big)^2}$

$\tt : \implies {\Big(k - 4\Big)}^2+{ \Big(2 - k\Big)^2}= 1+{9}$

$\tt : \implies k^2+16 - 8k+{4+k^2 - 4k}= 10.$

$\tt : \implies 2k^2- 12k + 20= 10.$

$\tt : \implies 2k^2- 12k + 10= 0.$

$\tt : \implies k^2- 6k + 5 = 0.$

$\tt : \implies k^2- 5k - k + 5 = 0.$

$\tt : \implies k(k- 5) - 1(k - 5) = 0.$

$\tt : \implies (k- 5) (k- 1) = 0.$

so,

$\tt \mapsto k - 5 = 0.$

$\tt \mapsto k = 5.$

Or,

$\tt \mapsto k - 1= 0.$

$\tt \mapsto k = 1.$

Therefore, the value of k is 5 and 1.

Answer 2

Answer:

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