Let the line segment joining the points A(2, -2) and B(3, 7) is divided by the line 2x+y-4 = 0 into ratio k:1. Using section formula ((m₁x₂+m₂x₁)/(m₁+m₂), (m₁y₂+m₂y₁)/(m₁+m₂)), the coordinates of point of intersection = ((3k+2)/(7k-2)/(k+1)) Since this point lies on the line 2x+y-4 = 0, so it will sRead more
Let the line segment joining the points A(2, -2) and B(3, 7) is divided by the line 2x+y-4 = 0 into ratio k:1.
Using section formula ((m₁x₂+m₂x₁)/(m₁+m₂), (m₁y₂+m₂y₁)/(m₁+m₂)), the coordinates of point of intersection = ((3k+2)/(7k-2)/(k+1))
Since this point lies on the line 2x+y-4 = 0, so it will satisfy the equar=tion of line. Therefore,
2((3k+2)/(k+1))+((7k-2)/(k+1))-4=0
⇒ (6k+4+7k-2-4k-4)/(k+1) = 0
⇒ 9k-2 = 0 ⇒ k = 2/9
Hence, the line segment joining the points A(2, -2) and B(3, 7) is divided by the line 2x+y-4 = 0 into ratio 2:9.
Since, points (x, y), (1.2) and (7,0) are collinear. So area of triangle formed by these points will be zero. Area of triangle = 1/2[x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)] ⇒ Area = 1/2[x(2-0) + 1(0-y)+ 7(y-2)] ⇒ 0 = 1/2[2x-y+7y-14] ⇒ 0 = [2x+6y-14] ⇒ x+3y-7 = 0 This is the video explanation of this questionRead more
Since, points (x, y), (1.2) and (7,0) are collinear. So area of triangle formed by these points will be zero.
Area of triangle = 1/2[x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)]
⇒ Area = 1/2[x(2-0) + 1(0-y)+ 7(y-2)]
⇒ 0 = 1/2[2x-y+7y-14]
⇒ 0 = [2x+6y-14]
⇒ x+3y-7 = 0
Let the centre of the circle passing through A(6, -6), B(3, -7) and C(3, 3) be O(x, y). OA = OB ⇒ √((x-6)²+(y+6)²) = √((x-3)²+(y+7)²) ⇒ √(x²+36-12x+y²+36+12y) = √(x²+9-6x+y²+49+14y) Squaring both the sides x²+36-12x+y²+36+12y = x²+9-6x+y²+49+14y ⇒ 72-12x+12y = 58-6x+14y ⇒ -6x-2y = 58-72 ⇒ -6x-2y = -Read more
Let the centre of the circle passing through A(6, -6), B(3, -7) and C(3, 3) be O(x, y).
OA = OB
⇒ √((x-6)²+(y+6)²) = √((x-3)²+(y+7)²)
⇒ √(x²+36-12x+y²+36+12y) = √(x²+9-6x+y²+49+14y)
Squaring both the sides
x²+36-12x+y²+36+12y = x²+9-6x+y²+49+14y
⇒ 72-12x+12y = 58-6x+14y
⇒ -6x-2y = 58-72
⇒ -6x-2y = -14
⇒ 3x+y = 7
⇒ y = 7-3x …(i)
Similarly, OA = OC [Radii od circle]
⇒ √((x-6)²+(y+6)²) = √((x-3)²+(y-3)²)
⇒ √(x²+36-12x+y²+36+12y+36+12y) = √(x²+9-6x+y²+9-6y)
Squaring both the sides
x²+36-12x+y²+36+12y = x²+9-6x+y²+9-6y
⇒ 72-12x+12y = 18-6x-6y
⇒ -6x+18y = 18-72
⇒ -6x+18y = -54
⇒ x-3y = 9
⇒ x-3(7-3x) = 9
⇒ x-21+9x = 9
10x = 30 ⇒ x = 3
Putting the value of x in equation (i), we get
y = 7-3(3) = -2
Hence, the centre of circle is O(3, -2).
The two opposite vertices of a square are (-1,2) and (3, 2). Let the two other vertices be B[x, y) and D(x₁, y₁) The sides of square are equal. Therefore, ∴ AB = BC ⇒ √((x+1)²(y-2)²) = √((x-3)²+(y-2)² ⇒ √(x²+1+2x+y²+4-4y) = √(x²+9-6x+y²+4-4y) Squaring both sides, we get x²+1+2x +y + 4 -4y = x²+9-6xRead more
The two opposite vertices of a square are (-1,2) and (3, 2). Let the two other vertices be B[x, y) and D(x₁, y₁)
The sides of square are equal. Therefore,
∴ AB = BC
⇒ √((x+1)²(y-2)²) = √((x-3)²+(y-2)²
⇒ √(x²+1+2x+y²+4-4y) = √(x²+9-6x+y²+4-4y)
Squaring both sides, we get
x²+1+2x +y + 4 -4y = x²+9-6x + y²+4-4y
⇒ x = 8 ⇒ x = 1
All the angles of square is 90°. Therefore, in AABC,
AB²+BC² = AC²
⇒ (√((x+1)²+(y-2)²))+(√((1-3)²+(y-2)²)) = (√(3+1)²+(2-2)²))
⇒ 4+y²+4-4y+4+y²-4y+4 = 16
⇒ 2y²+16-8y = 0
⇒ y(y-4) = 0
⇒ y = 0 or y = 4
We know that, the diagonals of square bisect each other. Therefore oordinates of mid-points of AC = Coordinates of mid-points of BD
⇒ ((-1+3)/2, (2+2)/2) = ((x+x₁)/2, (y+y₁)/2)
⇒ (1, 2) = ((1+x₁)/2, (y+y₁)/2)
On comparing, we get,
(1+x₁)/2 = 1
⇒ 1+x₁ = 2 ⇒ x₁ = 1
and (y+y₁)/2 ⇒ y+y₁ = 4
If y = 0, y₁ = 4,
If y = 4, y₁ = 0
Hence, the coordinates of other two vertices of square are (1, 0) and (1, 4).
Given that: AD/AB = AE/AC = 1/4 Let AD = x, therefore AB = 4x as AD/AB = 1/4 Hence, BD = AB-AD = 4x-x = 3x ⇒ AD/DB = 1/3 Similarly, AE/EC = 1/3 Hence, the points D and E divides the sides AB and AC respectively into 1:3. Therefore, Coordinates of point D = ((1×1+3×4)/(1+3), (1×5+3×6)/(1+3)) = (13/4,Read more
Given that: AD/AB = AE/AC = 1/4
Let AD = x, therefore AB = 4x as AD/AB = 1/4
Hence, BD = AB-AD = 4x-x = 3x
⇒ AD/DB = 1/3
Similarly,
AE/EC = 1/3
Hence, the points D and E divides the sides AB and AC respectively into 1:3.
Therefore,
Coordinates of point D = ((1×1+3×4)/(1+3), (1×5+3×6)/(1+3)) = (13/4, 23/4)
Coordinates of point E = ((1×7+3×4)/(1+3), (1×2+3×6)/(1+3)) = (19/4, 20/4)
Area of Triangle ADE
= 1/2[x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)]
= 1/2[4((23/4)-(20/4))+13/4((20/4)-6)+19/4(6-(23/4))]
= 1/2[3- (13/4)+(19/16)]
= 1/2[(48-52+19)/16]
= 15/32 square units
Area of triangle ACB,
= 1/2[x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)]
= 1/2[4(5-2)+1(2-6)+7(6-5)]
= 1/2[12-4+7]
= 15/2 square units
Now, (Area of triangle ADE)/(Area of triangle ABC) = (15/32)/(15/2) = 2/32 = 1/16
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Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, – 2) and B(3, 7).
Let the line segment joining the points A(2, -2) and B(3, 7) is divided by the line 2x+y-4 = 0 into ratio k:1. Using section formula ((m₁x₂+m₂x₁)/(m₁+m₂), (m₁y₂+m₂y₁)/(m₁+m₂)), the coordinates of point of intersection = ((3k+2)/(7k-2)/(k+1)) Since this point lies on the line 2x+y-4 = 0, so it will sRead more
Let the line segment joining the points A(2, -2) and B(3, 7) is divided by the line 2x+y-4 = 0 into ratio k:1.
Using section formula ((m₁x₂+m₂x₁)/(m₁+m₂), (m₁y₂+m₂y₁)/(m₁+m₂)), the coordinates of point of intersection = ((3k+2)/(7k-2)/(k+1))
Since this point lies on the line 2x+y-4 = 0, so it will satisfy the equar=tion of line. Therefore,
2((3k+2)/(k+1))+((7k-2)/(k+1))-4=0
⇒ (6k+4+7k-2-4k-4)/(k+1) = 0
⇒ 9k-2 = 0 ⇒ k = 2/9
Hence, the line segment joining the points A(2, -2) and B(3, 7) is divided by the line 2x+y-4 = 0 into ratio 2:9.
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See lessFind a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.
Since, points (x, y), (1.2) and (7,0) are collinear. So area of triangle formed by these points will be zero. Area of triangle = 1/2[x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)] ⇒ Area = 1/2[x(2-0) + 1(0-y)+ 7(y-2)] ⇒ 0 = 1/2[2x-y+7y-14] ⇒ 0 = [2x+6y-14] ⇒ x+3y-7 = 0 This is the video explanation of this questionRead more
Since, points (x, y), (1.2) and (7,0) are collinear. So area of triangle formed by these points will be zero.
Area of triangle = 1/2[x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)]
⇒ Area = 1/2[x(2-0) + 1(0-y)+ 7(y-2)]
⇒ 0 = 1/2[2x-y+7y-14]
⇒ 0 = [2x+6y-14]
⇒ x+3y-7 = 0
This is the video explanation of this question😎
See lessFind the center of a circle passing through the points (6, – 6), (3, – 7) and (3, 3).
Let the centre of the circle passing through A(6, -6), B(3, -7) and C(3, 3) be O(x, y). OA = OB ⇒ √((x-6)²+(y+6)²) = √((x-3)²+(y+7)²) ⇒ √(x²+36-12x+y²+36+12y) = √(x²+9-6x+y²+49+14y) Squaring both the sides x²+36-12x+y²+36+12y = x²+9-6x+y²+49+14y ⇒ 72-12x+12y = 58-6x+14y ⇒ -6x-2y = 58-72 ⇒ -6x-2y = -Read more
Let the centre of the circle passing through A(6, -6), B(3, -7) and C(3, 3) be O(x, y).
OA = OB
⇒ √((x-6)²+(y+6)²) = √((x-3)²+(y+7)²)
⇒ √(x²+36-12x+y²+36+12y) = √(x²+9-6x+y²+49+14y)
Squaring both the sides
x²+36-12x+y²+36+12y = x²+9-6x+y²+49+14y
⇒ 72-12x+12y = 58-6x+14y
⇒ -6x-2y = 58-72
⇒ -6x-2y = -14
⇒ 3x+y = 7
⇒ y = 7-3x …(i)
Similarly, OA = OC [Radii od circle]
⇒ √((x-6)²+(y+6)²) = √((x-3)²+(y-3)²)
⇒ √(x²+36-12x+y²+36+12y+36+12y) = √(x²+9-6x+y²+9-6y)
Squaring both the sides
x²+36-12x+y²+36+12y = x²+9-6x+y²+9-6y
⇒ 72-12x+12y = 18-6x-6y
⇒ -6x+18y = 18-72
⇒ -6x+18y = -54
⇒ x-3y = 9
⇒ x-3(7-3x) = 9
⇒ x-21+9x = 9
10x = 30 ⇒ x = 3
Putting the value of x in equation (i), we get
y = 7-3(3) = -2
Hence, the centre of circle is O(3, -2).
Video explanation is here 👇
See lessThe two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices.
The two opposite vertices of a square are (-1,2) and (3, 2). Let the two other vertices be B[x, y) and D(x₁, y₁) The sides of square are equal. Therefore, ∴ AB = BC ⇒ √((x+1)²(y-2)²) = √((x-3)²+(y-2)² ⇒ √(x²+1+2x+y²+4-4y) = √(x²+9-6x+y²+4-4y) Squaring both sides, we get x²+1+2x +y + 4 -4y = x²+9-6xRead more
The two opposite vertices of a square are (-1,2) and (3, 2). Let the two other vertices be B[x, y) and D(x₁, y₁)
The sides of square are equal. Therefore,
∴ AB = BC
⇒ √((x+1)²(y-2)²) = √((x-3)²+(y-2)²
⇒ √(x²+1+2x+y²+4-4y) = √(x²+9-6x+y²+4-4y)
Squaring both sides, we get
x²+1+2x +y + 4 -4y = x²+9-6x + y²+4-4y
⇒ x = 8 ⇒ x = 1
All the angles of square is 90°. Therefore, in AABC,
AB²+BC² = AC²
⇒ (√((x+1)²+(y-2)²))+(√((1-3)²+(y-2)²)) = (√(3+1)²+(2-2)²))
⇒ 4+y²+4-4y+4+y²-4y+4 = 16
⇒ 2y²+16-8y = 0
⇒ y(y-4) = 0
⇒ y = 0 or y = 4
We know that, the diagonals of square bisect each other. Therefore oordinates of mid-points of AC = Coordinates of mid-points of BD
⇒ ((-1+3)/2, (2+2)/2) = ((x+x₁)/2, (y+y₁)/2)
⇒ (1, 2) = ((1+x₁)/2, (y+y₁)/2)
On comparing, we get,
(1+x₁)/2 = 1
⇒ 1+x₁ = 2 ⇒ x₁ = 1
and (y+y₁)/2 ⇒ y+y₁ = 4
If y = 0, y₁ = 4,
If y = 4, y₁ = 0
Hence, the coordinates of other two vertices of square are (1, 0) and (1, 4).
Video Explanation of this question 👇
See lessThe vertices of a triangle ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that AD/AB = AE/AC= ¼. Calculate the area of the angle ADE and compare it with the area of angle ABC. (Recall Theorem 6.2 and Theorem 6.6).
Given that: AD/AB = AE/AC = 1/4 Let AD = x, therefore AB = 4x as AD/AB = 1/4 Hence, BD = AB-AD = 4x-x = 3x ⇒ AD/DB = 1/3 Similarly, AE/EC = 1/3 Hence, the points D and E divides the sides AB and AC respectively into 1:3. Therefore, Coordinates of point D = ((1×1+3×4)/(1+3), (1×5+3×6)/(1+3)) = (13/4,Read more
Given that: AD/AB = AE/AC = 1/4
Let AD = x, therefore AB = 4x as AD/AB = 1/4
Hence, BD = AB-AD = 4x-x = 3x
⇒ AD/DB = 1/3
Similarly,
AE/EC = 1/3
Hence, the points D and E divides the sides AB and AC respectively into 1:3.
Therefore,
Coordinates of point D = ((1×1+3×4)/(1+3), (1×5+3×6)/(1+3)) = (13/4, 23/4)
Coordinates of point E = ((1×7+3×4)/(1+3), (1×2+3×6)/(1+3)) = (19/4, 20/4)
Area of Triangle ADE
= 1/2[x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)]
= 1/2[4((23/4)-(20/4))+13/4((20/4)-6)+19/4(6-(23/4))]
= 1/2[3- (13/4)+(19/16)]
= 1/2[(48-52+19)/16]
= 15/32 square units
Area of triangle ACB,
= 1/2[x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)]
= 1/2[4(5-2)+1(2-6)+7(6-5)]
= 1/2[12-4+7]
= 15/2 square units
Now, (Area of triangle ADE)/(Area of triangle ABC) = (15/32)/(15/2) = 2/32 = 1/16
Here, it is the video explanation of this question✌️
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