Determinants Test

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Determinants Test - 41

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Weekly Quiz Competition

Question 1
1 / -0

Value of p for which the points (-5, 1), (1, p) and (4, -2) are collinear is

A
0

B
2

C
-1

D
None of these

Solution

Since the given points are collinear, they do not form a

triangle, which means area of the triangle is Zero.

Area

of a triangle with vertices $$({ x }_{ 1 },{ y }_{ 1 })$$ ; $$({ x }_{ 2 },{ y

}_{ 2 })$$ and $$({ x }_{ 3 },{ y }_{ 3 })$$ is $$ \left| \dfrac { {

x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{

3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right| $$

Hence, substituting the points $$({ x }_{ 1 },{ y }_{ 1 }) = (-5,1) $$ ; $$({ x }_{

2 },{ y }_{ 2 }) = (1,p) $$ and $$({ x }_{ 3 },{ y }_{ 3 }) = (4,-2)$$ in

the area formula, we get

$$ \left| \dfrac { -5(p+2) + 1(-2-1) + 4(1-p) } {2} \right| =

0 $$

$$ => -9p-9 = 0 $$

$$ =>p = -1$$
Question 2
1 / -0

If the points (-2, -5), (2, -2) and (8, a) are collinear then value of a will be:

A
$$\displaystyle \frac { 1 }{ 2 } $$

B
$$\displaystyle \frac { 3 }{ 2 } $$

C
$$\displaystyle \frac { -5 }{ 2 } $$

D
$$\displaystyle \frac { 5 }{ 2 } $$

Solution

Given points $$A(-2,-5),B(2,-2),C(8,a)$$ are collinear
Hence area of triangle formed by A,B,C is zero
$$Area =\dfrac{1}{2}\begin{vmatrix}x_{1}&y_{1}&1\\x_{2}&y_{2}&1\\x_{3}&y_{3}&1\end{vmatrix}$$
$$0=\dfrac{1}{2}\begin{vmatrix}-2&-5&1\\2&-2&1\\8&a&1\end{vmatrix}$$
$$0=-2(-2-a)+5(2-8)+1(2a+16)$$
$$4+2a-30+2a+16=0$$
$$4a=10\Rightarrow a=\dfrac{5}{2}$$
Question 3
1 / -0

If the points $$A(1, 2) ,\ O (0, 0)$$ and $$C (a, b)$$ are collinear then

A
$$a = b$$

B
$$a = 2b$$

C
$$2a = b$$

D
$$a = -b$$

Solution

$$\textbf{Step -1: Area of triangle in co-ordinate geometry.}$$
$$\text{Given points are }A(1, 2), O(0, 0), C(a, b)$$

$$\text{The points are said to be collinear if the area of }\triangle AOC = 0$$

$$\text{Area of }\triangle AOC = $$$$\dfrac{1}{2}[x_1(y_2 - y_3) + x_2(y_3-y_1)+x_3(y_1-y_2)]=0$$

$$\textbf{Step -2: Substituting the given points in the formula.}$$

$$\text{Area of }\triangle AOC = $$$$\dfrac{1}{2}[1(0-b) + 0(b-2)+a(2-0)]=0$$

$$\Rightarrow - b +2a = 0$$

$$\Rightarrow 2a = b$$

$$\textbf{Hence, the points A(1, 2) , O(0, 0) and C(a, b) are collinear when }$$$$\textbf{2a = b.}$$
Question 4
1 / -0

The points (-a ,-b), (0, 0) (a, b) and $$\displaystyle (a^{2},ab)$$ are

A
Collnear

B
Vertices of parallelogram

C
Vertices of a rectangle

D
None of these

Solution

Let $$ A (-a, -b), B (0,0), C(a,b) $$ and $$ D({a}^{2},ab) $$
Distance between two points $$ \left( { x }_{ 1 },{ y }_{ 1
} \right) $$ and $$ \left( { x }_{ 2 },{ y }_{ 2 } \right) $$ can be calculated
using the formula $$ \sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2
}+\left( { y }_{ 2 }-{ y }_{ 1 } \right) ^{ 2 } } $$
Distance between the points A$$

(-a,-b) $$ and B $$ (0,0) = \sqrt { \left( 0 + a \right) ^{ 2 }+\left( 0 + b

\right) ^{ 2 } } = \sqrt { {a}^{2} + {b}^{2} } $$
Distance between the points B$$

(0,0) $$ and C $$ (a,b) = \sqrt { \left( a-0 \right) ^{ 2 }+\left( b-0

\right) ^{ 2 } } = \sqrt { {a}^{2} + {b}^{2} } $$
Distance between the points C$$

(a,b) $$ and D $$ ({a}^{2}, ab) = \sqrt { \left( {a}^{2} - a \right) ^{ 2 }+\left(ab - b

\right) ^{ 2 } } = (a-1)\sqrt { {a}^{2} + {b}^{2} } $$
Distance between the points A$$

(-a,-b) $$ and D $$ ({a}^{2}, ab) = \sqrt { \left( {a}^{2} + a \right) ^{ 2 }+\left(ab + b

\right) ^{ 2 } } = (a +1)\sqrt { {a}^{2} + {b}^{2} } $$
Now. $$ AB+BC+CD = \sqrt { {a}^{2} + {b}^{2} } + \sqrt { {a}^{2} + {b}^{2} }+ (a-1)\sqrt { {a}^{2} + {b}^{2} } = (a +1)\sqrt { {a}^{2} + {b}^{2} } = AD $$
Hence, the points are collinear.
Question 5
1 / -0

The points (-a, -b), (0, 0), (a, b) and ($$\displaystyle a^{2}$$, ab) are

A
Collinear

B
Vertices of a parallelogram

C
Vertices of a rectangle

D
None of these

Solution

Let $$ A (-a, -b), B (0,0), C(a,b) $$ and $$ D({a}^{2},ab) $$

Distance between two points $$ \left( { x }_{ 1 },{ y }_{ 1} \right) $$ and $$ \left( { x }_{ 2 },{ y }_{ 2 } \right) $$ can be calculated using the formula $$ \sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2}+\left( { y }_{ 2 }-{ y }_{ 1 } \right) ^{ 2 } } $$

Distance between the points A$$(-a,-b) $$ and B $$ (0,0) = \sqrt { \left( 0 + a \right) ^{ 2 }+\left( 0 + b \right) ^{ 2 } } = \sqrt { {a}^{2} + {b}^{2} } $$

Distance between the points B$$ (0,0) $$ and C $$ (a,b) = \sqrt { \left( a-0 \right) ^{ 2 }+\left( b-0\right) ^{ 2 } } = \sqrt { {a}^{2} + {b}^{2} } $$

Distance between the points C$$

(a,b) $$ and D $$ ({a}^{2}, ab) = \sqrt { \left( {a}^{2} - a \right) ^{ 2 }+\left(ab - b

\right) ^{ 2 } } = (a-1)\sqrt { {a}^{2} + {b}^{2} } $$
Distance between the points A$$(-a,-b) $$ and D $$ ({a}^{2}, ab) = \sqrt { \left( {a}^{2} + a \right) ^{ 2 }+\left(ab + b\right) ^{ 2 } } = (a +1)\sqrt { {a}^{2} + {b}^{2} } $$
Now.

$$ AB+BC+CD = \sqrt { {a}^{2} + {b}^{2} } + \sqrt { {a}^{2} + {b}^{2}

}+ (a-1)\sqrt { {a}^{2} + {b}^{2} } = (a +1)\sqrt { {a}^{2} + {b}^{2}

} = AD $$
Hence, the points are collinear.
Question 6
1 / -0

The points (0, 8/3), (1, 3) and (82, 30) are the vertices of :

A
obtuse angled triangle

B
right angled triangle

C
isosceles triangle

D
None of these

Solution

Let, $$ P =(0, \dfrac {8}{3}),Q=(1,3),R=(82,30) $$

Now, m1 $$ = $$ slope of PQ $$ = \dfrac{3 \dfrac {8}{3}}{10} = \dfrac {1}{3} $$

m2 $$ = $$ slope of QR $$ = \dfrac {303}{821} = \dfrac {1}{3} $$

Since, $$ m1 = m2 $$, the three points P, Q, R are collinear
Question 7
1 / -0

If the points $$(k, 2k),\ ( 3k, 3k)$$ and $$(3, 1)$$ are collinear then the value of $$k$$ is

A
$$\displaystyle \frac{7}{9}$$

B
$$\displaystyle \frac{2}{3}$$

C
$$\displaystyle \frac{-2}{3}$$

D
$$\displaystyle \frac{-1}{3}$$

Solution

Since the given points are collinear, they do not form a triangle, which means area of the triangle is Zero
The area of the triangle with vertices $$({ x }_{ 1 },{ y }_{ 1 })$$ ; $$({ x }_{ 2 },{ y}_{ 2 })$$ and $$({ x }_{ 3 },{ y }_{ 3 })$$ is $$ \left| \frac { {x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right| $$
Hence, substituting the points $$({ x }_{ 1 },{ y }_{ 1 }) = (k,2k) $$ ; $$({ x}_{ 2 },{ y }_{ 2 }) = (3k,3k) $$ and $$({ x }_{ 3 },{ y }_{ 3 }) = (3,1)$$ in the area formula, we get $$ \left| \frac { k(3k-1)+3k(1-2k)+3(2k-3k) }{ 2 } \right| =0 $$
$$ => 3{k}^{2} -k + 3k-6{k}^{2} + 6k -9k = 0 $$
$$ =>-3{k}^{2} -k = 0 $$
$$ -k(3k+1) =0 $$
$$ k = 0 $$ or $$ k = -\frac{1}{3} $$
Question 8
1 / -0

The points (-a, -b), (0, 0), (a, b) and (a$$^2$$, ab) are

A
collinear

B
vertices of a rectangle

C
vertices of a parallelogram

D
None of these

Solution

Let the points be $$ A (-a, -b), O (0, 0), B (a, b) $$ and $$ C ({a}^{2}, ab) $$

Slope

of the line passing through points $$\left( { x }_{ 1 },{ y }_{ 1 } \right) $$

and $$\left( { x }_{ 2 },{ y }_{ 2 } \right)$$ $$ = $$ $$\dfrac { { y }_{ 2 }-{

y }_{ 1 } }{ { x }_{ 2 }-x_{ 1 } } $$

So, slope of the line joining OA $$ = \dfrac { -b-0 }{ -a-0 } = \dfrac {b}{a} $$
Slope of the line joining OB $$ = \dfrac { b-0 }{ a-0} = \dfrac {b}{a} $$
Slope of the line joining OC $$ = \dfrac { ab-0 }{{a}^{2}-0 } = \dfrac {b}{a} $$

Since slopes of the lines, OA, OB, OC are equal, they are all collinear.
Question 9
1 / -0

Find the value of P for which the points A(-1, 3), B(2, P) and C(5, -1) are collinear :

A
3

B
1

C
2

D
4

Solution

Since the given points are collinear, they do not form a

triangle, which means area of the triangle is Zero.

Area of a triangle with vertices $$({ x }_{ 1 },{ y }_{ 1 })$$ ; $$({ x }_{ 2 },{ y

}_{ 2 })$$ and $$({ x }_{ 3 },{ y }_{ 3 })$$ is $$ \left| \dfrac { {

x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{

3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right| $$
Hence, substituting the points $$({ x }_{ 1 },{ y }_{ 1 }) = (-1,3) $$ ; $$({ x

}_{ 2 },{ y }_{ 2 }) = (2,P) $$ and $$({ x }_{ 3 },{ y }_{ 3 }) = (5,-1)$$

in the area formula, we get

$$ \left| \dfrac { (-1)(P + 1) + 2(-1-3) + 5(3-P) }{ 2 } \right| =

0 $$

$$ => -P - 1 - 8 + 15 - 5P = 0 $$

$$ => 6P = 6 $$
$$ => P = 1 $$
Question 10
1 / -0

If the points $$(5, 1),\ (1, p)\ \& \ (4, 2)$$ are collinear then the value of p will be

A
$$1$$

B
$$5$$

C
$$2$$

D
$$-2$$

Solution

Area of a triangle with vertices $$({ x }_{ 1 },{ y }_{ 1})$$ ; $$({ x }_{ 2 },{ y }_{ 2 })$$ and $$({ x }_{ 3 },{ y }_{ 3})$$ is $$ \left| \dfrac { { x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2}({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right| $$
Since the given points are collinear, they do not form a triangle, which means area of the triangle is Zero.
Hence, substituting the points $$({ x }_{ 1 },{ y }_{ 1 }) = (5,1) $$ ; $$({ x}_{ 2 },{ y }_{ 2 }) = (1,P) $$ and $$({ x }_{ 3 },{ y }_{ 3 }) = (4,2)$$ in the area formula, we get $$ \left| \dfrac { 5(P-2)+1(2-1)+4(1-P) }{ 2 } \right| =0 $$
$$ => 5P-10+1+4-4P = 0 $$
$$ => P = 5 $$

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Determinants Test - 41

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Determinants Test - 41

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SHARING IS CARING

Question 1

0

2

-1

None of these

Solution

Question 2

$$\displaystyle \frac { 1 }{ 2 } $$

$$\displaystyle \frac { 3 }{ 2 } $$

$$\displaystyle \frac { -5 }{ 2 } $$

$$\displaystyle \frac { 5 }{ 2 } $$

Solution

Question 3

$$a = b$$

$$a = 2b$$

$$2a = b$$

$$a = -b$$

Solution

Question 4

Collnear

Vertices of parallelogram

Vertices of a rectangle

None of these

Solution

Question 5

Collinear

Vertices of a parallelogram

Vertices of a rectangle

None of these

Solution

Question 6

obtuse angled triangle

right angled triangle

isosceles triangle

None of these

Solution

Question 7

$$\displaystyle \frac{7}{9}$$

$$\displaystyle \frac{2}{3}$$

$$\displaystyle \frac{-2}{3}$$

$$\displaystyle \frac{-1}{3}$$

Solution

Question 8

collinear

vertices of a rectangle

vertices of a parallelogram

None of these

Solution

Question 9

3

1

2

4

Solution

Question 10

$$1$$

$$5$$

$$2$$

$$-2$$

Solution

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