Given points: A(5, -2), B(6,4) and C(7,-2) are vertices of triangle. Using distance formula: √((x₂-x₁)²+(y₂-y₁)²) AB = √((6-5)²+[4-(-2)]²)P = √(1+36) = √37 BC = √((7-6)²+(-2-4)²) = √(1+36) = √37 CA = √((5-7)²+[-2-(-2)]²) = √(4+0) = 2 Here, AB = BC ≠ AC Hence, the points (5,-2),. (6,4) and (7,-2) areRead more
Given points: A(5, -2), B(6,4) and C(7,-2) are vertices of triangle.
Using distance formula: √((x₂-x₁)²+(y₂-y₁)²)
AB = √((6-5)²+[4-(-2)]²)P = √(1+36) = √37
BC = √((7-6)²+(-2-4)²) = √(1+36) = √37
CA = √((5-7)²+[-2-(-2)]²) = √(4+0) = 2
Here, AB = BC ≠ AC
Hence, the points (5,-2),. (6,4) and (7,-2) are the vertices of an isosceles triangle.
(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.
Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.
Given points: A(5, -2), B(6,4) and C(7,-2) are vertices of triangle. Using distance formula: √((x₂-x₁)²+(y₂-y₁)²) AB = √((6-5)²+[4-(-2)]²)P = √(1+36) = √37 BC = √((7-6)²+(-2-4)²) = √(1+36) = √37 CA = √((5-7)²+[-2-(-2)]²) = √(4+0) = 2 Here, AB = BC ≠ AC Hence, the points (5,-2),. (6,4) and (7,-2) areRead more
Given points: A(5, -2), B(6,4) and C(7,-2) are vertices of triangle.
Using distance formula: √((x₂-x₁)²+(y₂-y₁)²)
AB = √((6-5)²+[4-(-2)]²)P = √(1+36) = √37
BC = √((7-6)²+(-2-4)²) = √(1+36) = √37
CA = √((5-7)²+[-2-(-2)]²) = √(4+0) = 2
Here, AB = BC ≠ AC
Hence, the points (5,-2),. (6,4) and (7,-2) are the vertices of an isosceles triangle.
Here is the video explanation 😃👇
See lessName 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 less