(i) Given points A(-1,-2), B(1, 0),C(-1, 2) and D(-3,0). AB = √([1-(-1)]²+[0-(-2)]²) = √(4+4) = √8 = 2√2 BC = √((-1-1)²+(2-0)²) = √(4+4) = √8 = 2√2 CD = √([-3-(-1)]²)+(0-2)²) = √(4+4) = √8 = 2√2 DA = √([-1-(-3)]²)+(-2-0)²) = √(4+4) √8 = 2√2 All the sides of quadrilateral are equal, so it may be a sqRead more
(i) Given points A(-1,-2), B(1, 0),C(-1, 2) and D(-3,0).
AB = √([1-(-1)]²+[0-(-2)]²) = √(4+4) = √8 = 2√2
BC = √((-1-1)²+(2-0)²) = √(4+4) = √8 = 2√2
CD = √([-3-(-1)]²)+(0-2)²) = √(4+4) = √8 = 2√2
DA = √([-1-(-3)]²)+(-2-0)²) = √(4+4) √8 = 2√2
All the sides of quadrilateral are equal, so it may be a square or rhombus on the basis of its diagonal.
AC = √([1-(-1)]²+[2-(-2)]²) = √(0+16) = 4
BD = √((-3-1)²+(0-0)²) = √(16+0) = 4
Here, AB = BC = CD = DA and AC = BD
Hence, ABCD is a square.
(ii) Given points: A(-3,5), B(3,1), C(0,3) and D(-1,-4).
AB = √([3-(-3)]²+(1-5)²) = √(36+6) = √52 = 2√13
BC = √((0-3)²+(3-1)²) = √(9+4) = √13
CD = √(1-0)²+(-4-3)²) = √(1+49) = √50 = 5√2
DA = √([-3-(-1)]²+[5-(-4)]²) = √(4+81) = √85
AC = √((-1-3)²+(-4-1)²) = √(16+25) = √41
BD = √((-1-3)²+(-4-1)²) = √(16+25) = √41
Here, AC+BC = AB, it means the point C lies on side AB or A,B,C are collinear.
Hence, the quadrilateral ABCD is not possible.
(iii) Given points: A(4,5), B(7,6), C(4,3) and D(1,2).
AB = √(7-4)²+(6-5)²) = √(9+1) = √10
BC = √((4-7)²+(3-6)²) = √(9+9) = √18
CD = √((1-4)²+(2-3)²) = √(9+1) = √10
DA = √((4-1)²+(5-2)]²) = √(9+9) = √18
The opposite sides of quadrilateral are equal. It may be a parallelogram or rectangle. It can be justified with the help of lengths of its diagonal.
AC = √((4-4)²+(3-5)²) = √(0+4) = 2
BD = √((1-7)²+(2-6)²) = √(36+16) = √52 = 2√13
Here, AB = CD, BC = AD and AC ≠ BD.
Hence, ABCD is a parallelogram.
Let P(x,0) be any point on x - axis, which is equidistant from A(2,-5) and B(-2,9). Therefore, PA = PB ⇒ √((2-x)²+(-5-0)²) = √((-2-x)² + (9-0)²) ⇒ √(4+x²-4x +25) = √(4+x²+4x+81) Squaring both the sides 4+x²-4x + 25 = 4+x² + 4x + 81 ⇒ -8x = 81- 25 = 56 ⇒ X = -56/8 = -7 Hence, P(-7,0) is the point onRead more
Let P(x,0) be any point on x – axis, which is equidistant from A(2,-5) and B(-2,9).
Therefore, PA = PB
⇒ √((2-x)²+(-5-0)²) = √((-2-x)² + (9-0)²)
⇒ √(4+x²-4x +25) = √(4+x²+4x+81)
Squaring both the sides
4+x²-4x + 25 = 4+x² + 4x + 81
⇒ -8x = 81- 25 = 56
⇒ X = -56/8 = -7
Hence, P(-7,0) is the point on the x-axis which is equidistant from (2,-5) and (-2,9).
To graphically represent a pair of linear equations in two variables, you first need to plot the equations on the xy-plane. It is done by substituting different values of x into the equations and finding the corresponding values of y. These pairs of x and y values can then be plotted as points on thRead more
To graphically represent a pair of linear equations in two variables, you first need to plot the equations on the xy-plane. It is done by substituting different values of x into the equations and finding the corresponding values of y. These pairs of x and y values can then be plotted as points on the xy-plane.
See the Video Solution 😁
An algebraic representation of a pair of linear equations in two variables allows us to solve the system of equations by using algebraic techniques, such as graphing, substitution, or elimination. See Here 👇
An algebraic representation of a pair of linear equations in two variables allows us to solve the system of equations by using algebraic techniques, such as graphing, substitution, or elimination.
Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (i) (– 1, – 2), (1, 0), (– 1, 2), (– 3, 0) (ii) (–3, 5), (3, 1), (0, 3), (–1, – 4) (iii) (4, 5), (7, 6), (4, 3), (1, 2)
(i) Given points A(-1,-2), B(1, 0),C(-1, 2) and D(-3,0). AB = √([1-(-1)]²+[0-(-2)]²) = √(4+4) = √8 = 2√2 BC = √((-1-1)²+(2-0)²) = √(4+4) = √8 = 2√2 CD = √([-3-(-1)]²)+(0-2)²) = √(4+4) = √8 = 2√2 DA = √([-1-(-3)]²)+(-2-0)²) = √(4+4) √8 = 2√2 All the sides of quadrilateral are equal, so it may be a sqRead more
(i) Given points A(-1,-2), B(1, 0),C(-1, 2) and D(-3,0).
AB = √([1-(-1)]²+[0-(-2)]²) = √(4+4) = √8 = 2√2
BC = √((-1-1)²+(2-0)²) = √(4+4) = √8 = 2√2
CD = √([-3-(-1)]²)+(0-2)²) = √(4+4) = √8 = 2√2
DA = √([-1-(-3)]²)+(-2-0)²) = √(4+4) √8 = 2√2
All the sides of quadrilateral are equal, so it may be a square or rhombus on the basis of its diagonal.
AC = √([1-(-1)]²+[2-(-2)]²) = √(0+16) = 4
BD = √((-3-1)²+(0-0)²) = √(16+0) = 4
Here, AB = BC = CD = DA and AC = BD
Hence, ABCD is a square.
(ii) Given points: A(-3,5), B(3,1), C(0,3) and D(-1,-4).
AB = √([3-(-3)]²+(1-5)²) = √(36+6) = √52 = 2√13
BC = √((0-3)²+(3-1)²) = √(9+4) = √13
CD = √(1-0)²+(-4-3)²) = √(1+49) = √50 = 5√2
DA = √([-3-(-1)]²+[5-(-4)]²) = √(4+81) = √85
AC = √((-1-3)²+(-4-1)²) = √(16+25) = √41
BD = √((-1-3)²+(-4-1)²) = √(16+25) = √41
Here, AC+BC = AB, it means the point C lies on side AB or A,B,C are collinear.
Hence, the quadrilateral ABCD is not possible.
(iii) Given points: A(4,5), B(7,6), C(4,3) and D(1,2).
AB = √(7-4)²+(6-5)²) = √(9+1) = √10
BC = √((4-7)²+(3-6)²) = √(9+9) = √18
CD = √((1-4)²+(2-3)²) = √(9+1) = √10
DA = √((4-1)²+(5-2)]²) = √(9+9) = √18
The opposite sides of quadrilateral are equal. It may be a parallelogram or rectangle. It can be justified with the help of lengths of its diagonal.
AC = √((4-4)²+(3-5)²) = √(0+4) = 2
BD = √((1-7)²+(2-6)²) = √(36+16) = √52 = 2√13
Here, AB = CD, BC = AD and AC ≠ BD.
Hence, ABCD is a parallelogram.
Here is the video explanation 😃👇
See lessFind the point on the x-axis which is equidistant from (2, –5) and (–2, 9).
Let P(x,0) be any point on x - axis, which is equidistant from A(2,-5) and B(-2,9). Therefore, PA = PB ⇒ √((2-x)²+(-5-0)²) = √((-2-x)² + (9-0)²) ⇒ √(4+x²-4x +25) = √(4+x²+4x+81) Squaring both the sides 4+x²-4x + 25 = 4+x² + 4x + 81 ⇒ -8x = 81- 25 = 56 ⇒ X = -56/8 = -7 Hence, P(-7,0) is the point onRead more
Let P(x,0) be any point on x – axis, which is equidistant from A(2,-5) and B(-2,9).
Therefore, PA = PB
⇒ √((2-x)²+(-5-0)²) = √((-2-x)² + (9-0)²)
⇒ √(4+x²-4x +25) = √(4+x²+4x+81)
Squaring both the sides
4+x²-4x + 25 = 4+x² + 4x + 81
⇒ -8x = 81- 25 = 56
⇒ X = -56/8 = -7
Hence, P(-7,0) is the point on the x-axis which is equidistant from (2,-5) and (-2,9).
See this video solution 😃👇
See lessThe coach of a cricket team buys 3 bats and 6 balls for Rupes 3900. Later, she buys another bat and 3 more balls of the same kind for Rupes 1300. Represent this situation algebraically and geometrically.
To graphically represent a pair of linear equations in two variables, you first need to plot the equations on the xy-plane. It is done by substituting different values of x into the equations and finding the corresponding values of y. These pairs of x and y values can then be plotted as points on thRead more
To graphically represent a pair of linear equations in two variables, you first need to plot the equations on the xy-plane. It is done by substituting different values of x into the equations and finding the corresponding values of y. These pairs of x and y values can then be plotted as points on the xy-plane.
See lessSee the Video Solution 😁
The cost of 2 kg of apples and 1kg of grapes on a day was found to be rupes 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is rupes 300. Represent the situation algebraically and geometrically
An algebraic representation of a pair of linear equations in two variables allows us to solve the system of equations by using algebraic techniques, such as graphing, substitution, or elimination. See Here 👇
An algebraic representation of a pair of linear equations in two variables allows us to solve the system of equations by using algebraic techniques, such as graphing, substitution, or elimination.
See Here 👇
See lessFind the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers. 26 and 91
26 and 91 26 = 2 × 13 91 = 7 × 13 HCF = 13 LCM = 2 × 7 × 13 = 182 Product of the two numbers = 26×91 = 2366 HCF × LCM = 13 × 182 = 2366 Hence, product of two numbers = HCF × LCM Here is the Video Solution 😎
26 and 91
26 = 2 × 13
91 = 7 × 13
HCF = 13
LCM = 2 × 7 × 13 = 182
Product of the two numbers = 26×91 = 2366
HCF × LCM = 13 × 182 = 2366
Hence, product of two numbers = HCF × LCM
Here is the Video Solution 😎
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