(i) (2,3),(-1,0),(2,-4) Vertices of triangle: A(2,3) ,B(-1,0) and C(2,-4) Using the formula for area of triangle ∆ = 1/2[x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)] Area of triangle ABC = 1/2[2{0-(-4)}+(-1){(-4)-3}+2(3-0)] = 1/2[8+7+6] = 21/2 = 10.5 square units (ii) (-5,-1),(3,-5),(5,2) Vertices of triangle: A(Read more
(i) (2,3),(-1,0),(2,-4)
Vertices of triangle: A(2,3) ,B(-1,0) and C(2,-4)
Using the formula for area of triangle ∆ = 1/2[x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)]
Area of triangle ABC = 1/2[2{0-(-4)}+(-1){(-4)-3}+2(3-0)]
= 1/2[8+7+6]
= 21/2 = 10.5 square units
(ii) (-5,-1),(3,-5),(5,2)
Vertices of triangle: A(-5,-1), B(3,-5) and C(5,2)
Using the formula for area of trangle ∆ = 1/2[x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)]
Area of triangle ABC = 1/2 [(-5)(-5-2)+3{3(2-(-1)}+5{-1(-5)}]
= 1/2[35+9+20]
= 64/2 = 32 square units
(i) A(7,-2), B(5, 1), C(3,k) Area of triangle formed by three collinear points is zero. Therefore, the area of triangle ABC = 0 ⇒ 1/2[7(1-k) +5{k- (-2)) +3(-2-1))] =0 ⇒ 7-7k+ 5k + 10 -9 = 0 ⇒ -2k = -8 ⇒ k = 4 (ii) P(8, 1), Q(k,-4), R(2,-5) Area of triangle formed by three collinear points is zero. TRead more
(i) A(7,-2), B(5, 1), C(3,k)
Area of triangle formed by three collinear points is zero.
Therefore, the area of triangle ABC = 0
⇒ 1/2[7(1-k) +5{k- (-2)) +3(-2-1))] =0
⇒ 7-7k+ 5k + 10 -9 = 0
⇒ -2k = -8
⇒ k = 4
(ii) P(8, 1), Q(k,-4), R(2,-5)
Area of triangle formed by three collinear points is zero.
Therefore, the area of triangle PQR = 0
⇒ 1/2[8{-4-(-5)+k(-5-1) + 2{1 -(-4)}] = 0
⇒ 8-6k+10 = 0
⇒ -6k = -18
⇒ k = 3
Coordinates of mid-points of AB = P((x₁+x₂)/2), (y₁+y₂)/2) = P((0=2)/2), (-1+1)/2) = P(1,0) Coordinates of mid-points of BC = R((2+0)/2, (1+3)/2) = R(1,2) Coordinates of mid-points of AC = Q((0+0)/2, (-1+3)/2) = Q(0,1) = 1/2[1(2-1)+1(1-0)+0(0-1)] = 1/2[2] = 1 square units Area of triangle ABC = 1/2[Read more
Coordinates of mid-points of AB = P((x₁+x₂)/2), (y₁+y₂)/2) = P((0=2)/2), (-1+1)/2) = P(1,0)
Coordinates of mid-points of BC = R((2+0)/2, (1+3)/2) = R(1,2)
Coordinates of mid-points of AC = Q((0+0)/2, (-1+3)/2) = Q(0,1)
= 1/2[1(2-1)+1(1-0)+0(0-1)] = 1/2[2] = 1 square units
Area of triangle ABC
= 1/2[0(1-3)+2{3-(-1)}+0(-1-1)] = 1/2[8] = 4 square units
Hence, (Area of triangle PQR)/(Area of triangle ABC) = 1/4
Area of triangle ABC = 1/2 [(-4){-5(-2)}+(-3){-2-(-2)}+3{-2-(-5)}] = 1/2 [12+0+9] = 1/2 [21] = 10.5 square units Area of triangle ACD = 1/2 [(-4)(-2-3)+3{3-(-2)}+2{-2-2(-2)}] = 1/2 [20+15+0] = 1/2[35] = 17.5 square units Area of quadrilateral ABCD = Area of triangle ABC +Area of triangle ACD ⇒ AreaRead more
Area of triangle ABC
= 1/2 [(-4){-5(-2)}+(-3){-2-(-2)}+3{-2-(-5)}]
= 1/2 [12+0+9]
= 1/2 [21] = 10.5 square units
Area of triangle ACD
= 1/2 [(-4)(-2-3)+3{3-(-2)}+2{-2-2(-2)}]
= 1/2 [20+15+0] = 1/2[35] = 17.5 square units
Area of quadrilateral ABCD = Area of triangle ABC +Area of triangle ACD
⇒ Area of quadrilateral ABCD = (10.5+17.5) square units = 28 square units
Coordinates pf mid-points of BC = D((x₁+y₂)/2, (y₁+y₂)/2) = D((3+5)/2, (-2+2)/2) = D(4, 0) Area of triangle ABD = 1/2[4(-2-0)+3{0-(-6)}+4{-6-(-2)}] = 1/2[-8+18-16] = 1/2[-6] = -3 = 3 square units Area of triangle ACD = 1/2 [4(2-0)+5{0-(-6)}+4(-6-2)] = 1/2 [8+30-32] = 1/2[6] = 3 square units ⇒ Area oRead more
Coordinates pf mid-points of BC = D((x₁+y₂)/2, (y₁+y₂)/2) = D((3+5)/2, (-2+2)/2) = D(4, 0)
Area of triangle ABD
= 1/2[4(-2-0)+3{0-(-6)}+4{-6-(-2)}]
= 1/2[-8+18-16] = 1/2[-6] = -3 = 3 square units
Area of triangle ACD
= 1/2 [4(2-0)+5{0-(-6)}+4(-6-2)]
= 1/2 [8+30-32] = 1/2[6] = 3 square units
⇒ Area of triangle ABD = Area of trinagle ACD = 3 square units
Hence, a medium of a triangle divides it into two triangles of equal areas.
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
Here, it is the video explanation of this question✌️
Find the area of the triangle whose vertices are : ( I ) (2, 3), (–1, 0), (2, – 4) ( ii) (–5, –1), (3, –5), (5, 2).
(i) (2,3),(-1,0),(2,-4) Vertices of triangle: A(2,3) ,B(-1,0) and C(2,-4) Using the formula for area of triangle ∆ = 1/2[x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)] Area of triangle ABC = 1/2[2{0-(-4)}+(-1){(-4)-3}+2(3-0)] = 1/2[8+7+6] = 21/2 = 10.5 square units (ii) (-5,-1),(3,-5),(5,2) Vertices of triangle: A(Read more
(i) (2,3),(-1,0),(2,-4)
Vertices of triangle: A(2,3) ,B(-1,0) and C(2,-4)
Using the formula for area of triangle ∆ = 1/2[x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)]
Area of triangle ABC = 1/2[2{0-(-4)}+(-1){(-4)-3}+2(3-0)]
= 1/2[8+7+6]
= 21/2 = 10.5 square units
(ii) (-5,-1),(3,-5),(5,2)
Vertices of triangle: A(-5,-1), B(3,-5) and C(5,2)
Using the formula for area of trangle ∆ = 1/2[x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)]
Area of triangle ABC = 1/2 [(-5)(-5-2)+3{3(2-(-1)}+5{-1(-5)}]
= 1/2[35+9+20]
= 64/2 = 32 square units
Here is the video explanation=>
See lessIn each of the following find the value of ‘k’, for which the points are collinear. (i) (7, –2), (5, 1), (3, k) (ii) (8, 1), (k, – 4), (2, –5)
(i) A(7,-2), B(5, 1), C(3,k) Area of triangle formed by three collinear points is zero. Therefore, the area of triangle ABC = 0 ⇒ 1/2[7(1-k) +5{k- (-2)) +3(-2-1))] =0 ⇒ 7-7k+ 5k + 10 -9 = 0 ⇒ -2k = -8 ⇒ k = 4 (ii) P(8, 1), Q(k,-4), R(2,-5) Area of triangle formed by three collinear points is zero. TRead more
(i) A(7,-2), B(5, 1), C(3,k)
Area of triangle formed by three collinear points is zero.
Therefore, the area of triangle ABC = 0
⇒ 1/2[7(1-k) +5{k- (-2)) +3(-2-1))] =0
⇒ 7-7k+ 5k + 10 -9 = 0
⇒ -2k = -8
⇒ k = 4
(ii) P(8, 1), Q(k,-4), R(2,-5)
Area of triangle formed by three collinear points is zero.
Therefore, the area of triangle PQR = 0
⇒ 1/2[8{-4-(-5)+k(-5-1) + 2{1 -(-4)}] = 0
⇒ 8-6k+10 = 0
⇒ -6k = -18
⇒ k = 3
See this Video explanation of this question😎
See lessFind the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, –1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.
Coordinates of mid-points of AB = P((x₁+x₂)/2), (y₁+y₂)/2) = P((0=2)/2), (-1+1)/2) = P(1,0) Coordinates of mid-points of BC = R((2+0)/2, (1+3)/2) = R(1,2) Coordinates of mid-points of AC = Q((0+0)/2, (-1+3)/2) = Q(0,1) = 1/2[1(2-1)+1(1-0)+0(0-1)] = 1/2[2] = 1 square units Area of triangle ABC = 1/2[Read more
Coordinates of mid-points of AB = P((x₁+x₂)/2), (y₁+y₂)/2) = P((0=2)/2), (-1+1)/2) = P(1,0)
Coordinates of mid-points of BC = R((2+0)/2, (1+3)/2) = R(1,2)
Coordinates of mid-points of AC = Q((0+0)/2, (-1+3)/2) = Q(0,1)
= 1/2[1(2-1)+1(1-0)+0(0-1)] = 1/2[2] = 1 square units
Area of triangle ABC
= 1/2[0(1-3)+2{3-(-1)}+0(-1-1)] = 1/2[8] = 4 square units
Hence, (Area of triangle PQR)/(Area of triangle ABC) = 1/4
See this video for better explanation✌️
See lessFind the area of the quadrilateral whose vertices, taken in order, are (– 4, – 2), (– 3, – 5), (3, – 2) and (2, 3).
Area of triangle ABC = 1/2 [(-4){-5(-2)}+(-3){-2-(-2)}+3{-2-(-5)}] = 1/2 [12+0+9] = 1/2 [21] = 10.5 square units Area of triangle ACD = 1/2 [(-4)(-2-3)+3{3-(-2)}+2{-2-2(-2)}] = 1/2 [20+15+0] = 1/2[35] = 17.5 square units Area of quadrilateral ABCD = Area of triangle ABC +Area of triangle ACD ⇒ AreaRead more
Area of triangle ABC
= 1/2 [(-4){-5(-2)}+(-3){-2-(-2)}+3{-2-(-5)}]
= 1/2 [12+0+9]
= 1/2 [21] = 10.5 square units
Area of triangle ACD
= 1/2 [(-4)(-2-3)+3{3-(-2)}+2{-2-2(-2)}]
= 1/2 [20+15+0] = 1/2[35] = 17.5 square units
Area of quadrilateral ABCD = Area of triangle ABC +Area of triangle ACD
⇒ Area of quadrilateral ABCD = (10.5+17.5) square units = 28 square units
See this Video Explanation😁
See lessYou have studied in Class IX, (Chapter 9, Example 3), that a median of a triangle divides it into two triangles of equal areas. Verify this result for engle ABC whose vertices are A(4, – 6), B(3, –2) and C(5, 2).
Coordinates pf mid-points of BC = D((x₁+y₂)/2, (y₁+y₂)/2) = D((3+5)/2, (-2+2)/2) = D(4, 0) Area of triangle ABD = 1/2[4(-2-0)+3{0-(-6)}+4{-6-(-2)}] = 1/2[-8+18-16] = 1/2[-6] = -3 = 3 square units Area of triangle ACD = 1/2 [4(2-0)+5{0-(-6)}+4(-6-2)] = 1/2 [8+30-32] = 1/2[6] = 3 square units ⇒ Area oRead more
Coordinates pf mid-points of BC = D((x₁+y₂)/2, (y₁+y₂)/2) = D((3+5)/2, (-2+2)/2) = D(4, 0)
Area of triangle ABD
= 1/2[4(-2-0)+3{0-(-6)}+4{-6-(-2)}]
= 1/2[-8+18-16] = 1/2[-6] = -3 = 3 square units
Area of triangle ACD
= 1/2 [4(2-0)+5{0-(-6)}+4(-6-2)]
= 1/2 [8+30-32] = 1/2[6] = 3 square units
⇒ Area of triangle ABD = Area of trinagle ACD = 3 square units
Hence, a medium of a triangle divides it into two triangles of equal areas.
See the video explanation here✌️
See lessDetermine 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.
See here for video explanation of this question😎
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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