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If (x₁, y₁), (x₂ , y₂), (x_3, y₃) (x₁, y₁) (x₂, y₂), (x₃, y₃) are the vertices of a triangle then its area is given by 

Area = [1/2 (x₁(y₂ −y₃)+x₂ (y₃ −y₁ )+x₃ (y₁ −y₂ ))]

Area = \(\frac{1}2\)[(2(0-6)+6(6-0)+4(0-0)]

⇔ \(\frac{1}2\)[-12+36+0]

⇒ 12 sq. units

The ordinate of any point on x-axis is always zero. This means that this point hasn’t covered at any distance on y-axis.

The abscissa of any point on y-axis is always zero. This means that this point hasn’t covered at any distance on x-axis.

The perpendicular distance of any point from y-axis is always equal to the value of abscissa

As we know in the fourth coordinate abscissa is positive and ordinate is negative.

We know that abscissa is always positive in first and fourth coordinate and ordinate is always positive in first and second coordinate.

As we know that abscissa is negative in second and third coordinate and ordinate is positive in first and second coordinate. Therefore the given point -3, 2 lies in second coordinate.

The perpendicular distance of any point from x-axis is always equal to the value of ordinate.

Two points having same abscissa but different ordinate always amke a line which is parallel to y-axis.

Correct option is C) \(\left(\cfrac{m_1x_2+m_2x_1}{m_1+m_2},\cfrac{m_1 y_2+m_2y_1}{m_1+m_2}\right)\)

Correct option is (D) Q₃

\(\because\) \((x_1,y_1)\) lies in \(Q_2\)

\(\therefore\) \(x_1<0\) & \(y_1>0\)

Also given that \((x_2,y_2)\) lies in \(Q_3.\)

\(\therefore\) \(x_2<0\), \(y_2<0\)

\(\because\) \(x_1<0\) & \(y_2<0\)

\(\therefore\) \((x_1,y_2)\) lies in \(Q_3\)

Correct option is  D) Q₃

Correct option is (D) 3abc

Centroid of triangle whose vertices are \(A(a,b),B(b,c)\;\&\;C(c,a)\) is \((\frac{a+b+c}3,\frac{b+c+a}3)=(0,0)\)         (Given)

\(\Rightarrow\) \(\frac{a+b+c}3=0\)                (By comparing)

\(\Rightarrow\) \(a+b+c=0\)

\(\because\) If \(a+b+c=0,\) then \(a^3+b^3+c^3=3abc\)

Correct option is D) 3abc

Correct option is (C) 2

\(P=\{3x|x\in N,x\leq50\}\)

\(\Rightarrow P=\{3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48\}\)

And \(Q=\{5x|x\in N,x\leq30\}\)

\(\Rightarrow Q=\{5,10,15,20,25,30\}\)

\(\therefore\) \(P\cap Q=\{15,30\}\)

\(\therefore\) \(n(P\cap Q)=2\)

Correct option is C) 2

Correct answer is (b) 4 units

The y-coordinate the distance of the point from the x - axis

Here, the y - coordinate is 4.

The given points A(−2, 1), B(a, b) and C(4, −1) are collinear. 

Area of the triangle having vertices (x₁,y₁), (x₂,y₂) and (x₃,y₃) 

= \(\frac{1}2\) |x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)| 

Given that area of ∆ABC = 0 

∴ −2[b − ( − 1)] + a( − 1 – 1 ) + 4 ( 1 –b ) = 0 

-2b – 2 − 2a + 4 − 4b = 0 

− 2a − 6b = − 2 

a + 3b = 1 …(1) 

Also it is given that a – b = 1 …(2) 

Solving 1 and 2 simultaneously, 

b + 1 + 3b = 1 

4b = 0 

∴ b = 0 

∴ a =1 

Hence, 

the values of a and b are 1 and 0.

Given vertices of a quadrilateral ABCD are A(−3, 5), B(−2, −7), C(1, −8) and D(6, 3) 

Area of the quadrilateral ABCD = Area of ∆ABC + Area of ∆ACD 

Area of the triangle having vertices (x₁,y₁), (x₂,y₂) and (x₃,y₃) 

= \(\frac{1}2\) |x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)| 

Area of ∆ABC = \(\frac{1}2\) | − 3[ − 7 − ( − 8 )] + ( − 2) ( − 8 – 5 ) + 1 [ 5 − (− 7) ] | 

= \(\frac{1}2\) | − 3 + 26 + 12 | 

= \(\frac{35}2\) sq. units 

Area of ∆ACD = \(\frac{1}2\) |− 3( − 8 – 3 ) + 1( 3 – 5 ) + 6[ 5 − ( − 8 ) ] | 

= \(\frac{1}2\) | 33 – 2 + 78 | 

= \(\frac{109}2\) sq. units 

Area of the quadrilateral ABCD = \(\frac{35}2\) + \(\frac{109}2\) = 72 sq. units 

∴Hence, 

the area of the quadrilateral is 72 sq. units.

Let P(−5,−3); Q(−4,−6); R(2,−3) and S(1,2) be the vertices of quadrilateral PQRS. 

Area of the quadrilateral PQRS = Area of ∆PQR + Area of ∆PSR 

Area of the triangle having vertices (x₁,y₁), (x₂,y₂) and (x₃,y₃) 

= \(\frac{1}2\) |x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)| 

Area of ∆PQR = \(\frac{1}2\) |− 5( − 6 + 3 ) − 4( − 3 + 3 ) + 2( −3 + 6) | 

= \(\frac{1}2\) |15 + 0 + 6| 

= \(\frac{21}2\) sq. units 

Area of ∆PSR = \(\frac{1}2\) | − 5( 2 + 3 ) + 1( − 3 + 3 ) + 2( − 3 − 2) | 

= \(\frac{1}2\) | − 25 + 0 – 10 | 

= \(\frac{25}2\) sq. units 

Area of the quadrilateral PQRS = \(\frac{21}2\)+\(\frac{25}2\) = 28 sq. units 

∴Hence, 

the area of the quadrilateral is 28 sq. units. 

Let the co-ordinates of P and R be (a,b) and (c,d) and coordinates of Q are (3, 2) 

By midpoint formula. 

x = \(\frac{x_1+x_2}2\) , y = \(\frac{y_1+y_2}2\)

(2 , - 1) is the mid-point of PQ. 

∴ 2 = \(\frac{3+a}2\) and -1 = \(\frac{2+b}2\)

∴ a = 1 and b = -4 

∴ Coordinates of P are (1, -4) 

(1 , 2) is the mid-point of QR. 

∴ 1 = \(\frac{3+c}2\) and 2 = \(\frac{2+d}2\)

∴ c = -1 and d = 2 

∴ Coordinates of P are (-1, 2) 

Area of the triangle having vertices (x₁,y₁), (x₂,y₂) and (x₃,y₃) 

= \(\frac{1}2\) |x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)| 

Area of ∆PQR = \(\frac{1}2\) | 3( − 4 – 2 ) + 2( − 1 – 1) + 1( 2 − 4) | 

= \(\frac{1}2\) | − 18 − 4 − 2 | 

= 12 sq. units 

Hence the area of ∆PQR is 12 sq. units