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NCERT10th MathematicsExercise 7.4Topic: 7 Coordinate geometry
EXERCISE 7.4
Q 1. 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).
Explanation:
1) The point P(x, y) divides the segment joining the points A(x1, y1) and B(x2, y2) in
the ratio m:n, so we have,
x = (mx2 + nx1)/(m + n)
2) In general P(x, y) = ((mx2 + nx1)/(m + n), (my2 + ny1)/(m + n))y = (my2 + ny1)/(m + n)
Solution:
1) Here line 2x + y – 4 = 0 divides line segment AB with A(2, -2) and B(3, 7)
in the ratio k:1.
2) So, by section formula,
x = (3k + 2)/(k + 1) and y = (7k – 2)/(k + 1) ---------- equation 1
3) Put the values of x and y from equation 1 in 2x + y – 4 = 0, we get,
2x + y – 4 = 0
2[(3k + 2)/(k + 1)] + [(7k – 2)/(k + 1)] – 4 = 0
2[(3k + 2)] + [(7k – 2)] – 4(k + 1) = 0
6k + 4 + 7k – 2 – 4k – 4 = 0
6k + 7k – 4k – 2 = 0
6k + 3k – 2 = 0
9k = 2
k = 2/9
4) So the ratio is 2:9.
Q 2. Find a relation between x and y if the points (x, y), (1, 2), and (7, 0) are
collinear.
Explanation:
1) Let ABC be any triangle whose vertices are A(x1, y1), B(x2, y2) and C(x3, y3).
Area of a triangle = (1/2)[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
2) When the points are collinear, then the area of the triangle formed by these
points is always zero.
Solution:
1) Let us name these coordinates as A(x, y), B(1, 2) and C(7, 0), so here
a) x1 = x
b) y1 = yc) x2 = 1
d) y2 = 2
e) x3 = 73) We know that the area of the triangle ABC is as given below,
f) y3 = 0
Area of a triangle = (1/2)[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
Area of a triangle = (1/2)[(x)(2 – 0) + (1)(0 – y) + (7)((y) – 2)]
Area of a triangle = (1/2)[2x – y + (7)(y – 2)]
Area of a triangle = (1/2)[2x – y + 7y – 14]
Area of a triangle = (1/2)[2x + 6y – 14]
Area of a triangle = [x + 3y – 7] --------- equation 1
4) As the points are collinear, then the area of the triangle formed by these
points is always zero. So, from equation 1 we have,
x + 3y – 7 = 0
5) So the relation between x and y is x + 3y – 7 = 0.
Q 3. Find the center of a circle passing through the
points (6, – 6), (3, – 7), and (3, 3).
Explanation:
1) The distance between A(x1, y1) and B(x2, y2) can be calculated as:
(AB) = √[(x1 – x2)2 + (y1 – y2)2]
Solution:
1) O(x, y) is the centre of a circle.2) Points A(3, 3), B(3, - 7), and C(6, - 6) are the points on the circle.3) So, all the radii are the same, i.e. OA = OB = OC ---------- equation 14) First we will find OA, where O(x, y) and A(3, 3),a) x1 = x
a) y1 = y
a) x2 = 3
a) y2 = 3
5) We know that:(OA) = √[(x1 – x2)2 + (y1 – y2)2]
(OA) = √[(x – 3)2 + (y – 3)2] ---------- equation 2
6) Now we will find OB, where O(x, y) and B(3, - 7),a) x1 = x
a) y1 = y
a) x2 = 3
a) y2 = – 7
7) We know that:(OB) = √[(x1 – x2)2 + (y1 – y2)2]
(OB) = √[(x – 3)2 + (y – (– 7)2]
1) O(x, y) is the centre of a circle.
2) Points A(3, 3), B(3, - 7), and C(6, - 6) are the points on the circle.
3) So, all the radii are the same, i.e. OA = OB = OC ---------- equation 1
4) First we will find OA, where O(x, y) and A(3, 3),
a) x1 = x
a) y1 = y
a) x2 = 3
a) y2 = 3
5) We know that:
(OA) = √[(x1 – x2)2 + (y1 – y2)2]
(OA) = √[(x – 3)2 + (y – 3)2] ---------- equation 2
6) Now we will find OB, where O(x, y) and B(3, - 7),
a) x1 = x
a) y1 = y
a) x2 = 3
a) y2 = – 7
7) We know that:
(OB) = √[(x1 – x2)2 + (y1 – y2)2]
(OB) = √[(x – 3)2 + (y – (– 7)2]
(OB) = √[(x – 3)2 + (y + 7)2] ---------- equation 3
8) First we will find OC, where O(x, y) and C(6, - 6),
a) x1 = x
a) y1 = y
a) x2 = 6
a) y2 = - 6
9) We know that:
(OC) = √[(x1 – x2)2 + (y1 – y2)2]
(OC) = √[(x – 6)2 + (y – (– 6))2]
(OC) = √[(x – 6)2 + (y + 6)2] ---------- equation 4
10) from equations 1, 2, and 3, we have,
OA = OB
√[(x – 3)2 + (y – 3)2] = √[(x – 3)2 + (y + 7)2]
[(x – 3)2 + (y – 3)2] = [(x – 3)2 + (y + 7)2]
[(y + 7)2 – (y – 3)2] = [(x – 3)2 – (x – 3)2]
[(y + 7)2 – (y – 3)2] = 0
[(y + 7) – (y – 3)] [(y + 7) + (y – 3)] = 0
[7 + 3] [(2y + 4)] = 0
2[10] [(y + 2)] = 0
(y + 2) = 0
y = – 2 ---------- equation 5
11) from equations 1, 2, and 3, we have,
OA = OC
√[(x – 3)2 + (y – 3)2] = √[(x – 6)2 + (y + 6))2]
[(x – 3)2 + (y – 3)2] = [(x – 6)2 + (y + 6)2]
[(x – 3)2 – (x – 6)2] = [(y + 6)2 – (y – 3)2]
[(x – 3) – (x – 6)] [(x – 3) + (x – 6)] = [(y + 6) – (y – 3)] [(y + 6) + (y – 3)]
[x – 3 – x + 6)] [x – 3 + x – 6] = [y + 6 – y + 3] [y + 6 + y – 3)]
[3] [2x – 9] = [9] [2y + 3)]
[2x – 9] = [3] [2y + 3)]
2x – 9 = 6y + 9
2x – 6y = 18
x – 3y = 9 ---------- equation 6
12) From equations 5 and 6, we have,
x – 3y = 9
x – 3 (– 2) = 9
x + 6 = 9
x = 9 – 6
x = 3
13) So the coordinates of the center are O(3, – 2)
Q 4. The two opposite vertices of a square are (–1, 2) and (3, 2). Find the
coordinates of the other two vertices.
Solution:
1) According to the problem, here B(x1, y1), and A(- 1, 2), so,a) x1 = x1
a) y1 = y1
a) x2 = - 1
a) y2 = 2
2) We know that:(BA) = √[(x1 – x2)2 + (y1 – y2)2]
(BA) = √[(x1 – (– 1))2 + (y1 – 2)2]
(BA) = √[(x1 + 1)2 + (y1 – 2)2] ----------- equation 1
3) Now we will find BC with B(x1, y1), and C(3, 2), so,a) x1 = x1
a) y1 = y1
a) x2 = 3
a) y2 = 2
4) Using the distance formula, we have:(BC) = √[(x1 – x2)2 + (y1 – y2)2]
(BC) = √[(x1 – 3)2 + (y1 – 2)2] ----------- equation 2
5) As BA and BC are the sides of the square,
1) According to the problem, here B(x1, y1), and A(- 1, 2), so,
a) x1 = x1
a) y1 = y1
a) x2 = - 1
a) y2 = 2
2) We know that:
(BA) = √[(x1 – x2)2 + (y1 – y2)2]
(BA) = √[(x1 – (– 1))2 + (y1 – 2)2]
(BA) = √[(x1 + 1)2 + (y1 – 2)2] ----------- equation 1
3) Now we will find BC with B(x1, y1), and C(3, 2), so,
a) x1 = x1
a) y1 = y1
a) x2 = 3
a) y2 = 2
4) Using the distance formula, we have:
(BC) = √[(x1 – x2)2 + (y1 – y2)2]
(BC) = √[(x1 – 3)2 + (y1 – 2)2] ----------- equation 2
5) As BA and BC are the sides of the square,
BA = BC ----------- equation 3
6) From equations 1, 2, and 3, we have,
BA = BC
√[(x1 + 1)2 + (y1 – 2)2] = √[(x1 – 3)2 + (y1 – 2)2]
[(x1 + 1)2 + (y1 – 2)2] = [(x1 – 3)2 + (y1 – 2)2]
(x1 + 1)2 = (x1 – 3)2
x12 + 2x1 + 1 = x12 – 6x1 + 9
2x1 + 1 = – 6x1 + 97) So here, the coordinates of the point B are B(1, y1)
2x1 + 6x1 = 9 – 1
2x1 + 6x1 = 8
8x1 = 8
x1 = 8/8
x1 = 1 ----------- equation 3
8) In ∆ ABC, using Pythagoras' theorem, and from equations 1, 2, and 3,
we have, A(– 1, 2), B(1, y1), C(3, 2)
(BA)2 + (BC)2 = (AC)2(BA)2 + (BC)2 = (AC)2[(1 – (– 1))2 + (y1 – 2)2] + [(1 – 3)2 + (y1 – 2)2] = [(– 1 – 3)2 + (2 – 2)2]
[(1 + 1))2 + (y1 – 2)2] + [(– 2)2 + (y1 – 2)2] = [(– 4)2 + (0)2]
[(2)2 + (y1 – 2)2] + [(– 2)2 + (y1 – 2)2] = [(– 4)2]
[4 + (y1 – 2)2] + [4 + (y1 – 2)2] = [16]
8 + 2(y1 – 2)2 = 16
2(y1 – 2)2 = 16 – 8
2(y1 – 2)2 = 8
(y1 – 2)2 = 4
y1 – 2 = ± 2y1 = 2 ± 2
y1 = 2 + 2 or y1 = 2 – 2
y1 = 4 or y1 = 0
9) So the coordinates of B will be B(1, 4) or B(1, 0).
10) Now we can find the coordinates of point D(x2, y2), using the midpoint form.
11) Point O(x, y) is the midpoint of segment AC with A(– 1, 2) and C(3, 2), so
O(x, y) = ((– 1 + 3)/2, (2 + 2)/2)
O(x, y) = ((2)/2, (4)/2)
O(x, y) = (1, 2)
12) Case 1: Taking B(1, 4)
Point O(1, 2) is the midpoint of segment BD with B(1, 4) and D(x2, y2), so
O(1, 2) = ((1 + x2)/2, (4 + y2)/2)(1 + x2)/2 = 1, and (4 + y2)/2 = 2(1 + x2) = 2, and (4 + y2) = 4x2 = 2 – 1, and y2 = 4 – 4x2 = 1, and y2 = 0
13) So the coordinates of D will be D(1, 0)
14) Case 2: Taking B(1, 0)
Point O(1, 2) is the midpoint of segment BD with B(1, 0) and D(x2, y2), so
O(1, 2) = ((1 + x2)/2, (0 + y2)/2)(1 + x2)/2 = 1, and (0 + y2)/2 = 2(1 + x2) = 2, and (0 + y2) = 4x2 = 2 – 1, and y2 = 4 – 0x2 = 1, and y2 = 4
15) So the coordinates of D will be D(1, 4)
Q 5. The Class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1m from each other. There is a triangular grassy lawn in the plot as shown in the following Fig. The students are to sow seeds of flowering plants on the remaining area of the plot.
(i) Taking A as origin, find the coordinates of the vertices of the triangle.
(ii) What will be the coordinates of the vertices of ∆ PQR if C is the origin? Also calculate the areas of the triangles in these cases. What do you observe?
Solution:
1) By taking A as a center, from the figure, the coordinates of P, Q, and R will be:
P(4, 6), Q(3, 2), and R(6, 5)
2) According to the problem,a) x1 = 4
b) y1 = 6c) x2 = 3
d) y2 = 2
e) x3 = 6
f) y3 = 5
3) We know that the area of the triangle PQR is as given below,
Area of ∆ PQR = (1/2)[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
Area of ∆ PQR = (1/2)[(4)(2 – 5) + (3)(5 – 6) + (6)(6 – 2)]
Area of ∆ PQR = (1/2)[(4)(– 3) + (3)(– 1) + (6)(4)]
Area of ∆ PQR = (1/2)[(– 12) + (– 3) + (24)]
Area of ∆ PQR = (1/2)[– 15 + 24]
Area of ∆ PQR = (1/2)(9)
Area of ∆ PQR = 9/2 square units.------- equation 1
2) According to the problem,
a) x1 = 4
b) y1 = 6c) x2 = 3
d) y2 = 2
e) x3 = 63) We know that the area of the triangle PQR is as given below,
f) y3 = 5
Area of ∆ PQR = (1/2)[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
Area of ∆ PQR = (1/2)[(4)(2 – 5) + (3)(5 – 6) + (6)(6 – 2)]
Area of ∆ PQR = (1/2)[(4)(– 3) + (3)(– 1) + (6)(4)]
Area of ∆ PQR = (1/2)[(– 12) + (– 3) + (24)]
Area of ∆ PQR = (1/2)[– 15 + 24]
Area of ∆ PQR = (1/2)(9)
Area of ∆ PQR = 9/2 square units.------- equation 1
4) By taking C as a center, from the figure, the coordinates of P, Q, and R will be:
P(12, 2), Q(13, 6), and R(10, 3)
5) According to the problem,a) x1 = 12
b) y1 = 2c) x2 = 13
d) y2 = 6
e) x3 = 10
f) y3 = 3
6) We know that the area of the triangle PQR is as given below,
Area of ∆ PQR = (1/2)[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
Area of ∆ PQR = (1/2)[(12)(6 – 3) + (13)(3 – 2) + (10)(2 – 6)]
Area of ∆ PQR = (1/2)[(12)(3) + (13)(1) + (10)(– 4)]
Area of ∆ PQR = (1/2)[(36) + (13) + (– 40)]
Area of ∆ PQR = (1/2)[49 – 40]
Area of ∆ PQR = (1/2)(9)
Area of ∆ PQR = 9/2 square units.------- equation 2
7) The area of a triangle PQR will be the same even if it is located anywhere on the
a) x1 = 12
b) y1 = 2c) x2 = 13
d) y2 = 6
e) x3 = 106) We know that the area of the triangle PQR is as given below,
f) y3 = 3
Area of ∆ PQR = (1/2)[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
Area of ∆ PQR = (1/2)[(12)(6 – 3) + (13)(3 – 2) + (10)(2 – 6)]
Area of ∆ PQR = (1/2)[(12)(3) + (13)(1) + (10)(– 4)]
Area of ∆ PQR = (1/2)[(36) + (13) + (– 40)]
Area of ∆ PQR = (1/2)[49 – 40]
Area of ∆ PQR = (1/2)(9)
Area of ∆ PQR = 9/2 square units.------- equation 2
7) The area of a triangle PQR will be the same even if it is located anywhere on the
graph.
Q 6. The vertices of a ∆ 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 = 1/4. Calculate the area of the ∆ ADE and compare it with the area of ∆ ABC. (Recall Theorem 6.2 and Theorem 6.6).
Solution:
1) First we will find the coordinates of point D.2) As, AD/AB = 1/4, so AD/DB = 1/3.3) Here, point D(x, y) divides the segment AB with A(4, 6) and B(1, 5) in the ratio 1:3
3) Here, point D(x, y) divides the segment AB with A(4, 6) and B(1, 5) in the ratio 1:3
we now, will find x coordinate
x = (1 (1) + 3 (4))/(1 + 3)
x = (1 + 12)/(4)
x = (13)/(4)
x = 13/4 --------- equation 1
we now, will find y coordinate
y = (1 (5) + 3 (6))/(1 + 3)
y = (5 + 18)/(4)
y = (23)/(4)
y = 23/4 --------- equation 2
4) From equations 1 and 2, x = 13/4 and y = 23/4. So, the coordinates of D are D(13/4, 23/4).
we now, will find x coordinate
x = (1 (1) + 3 (4))/(1 + 3)
x = (1 + 12)/(4)
x = (13)/(4)
x = 13/4 --------- equation 1
y = (1 (5) + 3 (6))/(1 + 3)
y = (5 + 18)/(4)
y = (23)/(4)
y = 23/4 --------- equation 2
5) Now we will find the coordinates of point E.6) As, AE/AC = 1/4, so AE/EC = 1/3.7) Point E(x, y) divides the segment AC with A(4, 6) and C(7, 2) in the ratio 1:3
7) Point E(x, y) divides the segment AC with A(4, 6) and C(7, 2) in the ratio 1:3
we now, will find x coordinate
x = (1 (7) + 3 (4))/(1 + 3)
x = (7 + 12)/(4)
x = (19)/(4)
x = 19/4 --------- equation 3
we now, will find y coordinate
y = (1 (2) + 3 (6))/(1 + 3)
y = (2 + 18)/(4)
y = (20)/(4)
y = 5 --------- equation 4
8) From equations 3 and 4, x = 19/4 and y = 5. So, the coordinates of E are E(19/4, 5).
we now, will find x coordinate
x = (1 (7) + 3 (4))/(1 + 3)
x = (7 + 12)/(4)
x = (19)/(4)
x = 19/4 --------- equation 3
y = (1 (2) + 3 (6))/(1 + 3)
y = (2 + 18)/(4)
y = (20)/(4)
y = 5 --------- equation 4
9) Now we will find the area of ∆ ADE with A(4, 6), D(13/4, 23/4), and E(19/4, 5).10) According to the problem,a) x1 = 4
b) y1 = 6c) x2 = 13/4
d) y2 = 23/4
e) x3 = 19/4
f) y3 = 5
11) We know that the area of the triangle ADE is as given below,
Area of ∆ ADE = (1/2)[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
Area of ∆ ADE = (1/2)[(4)((23/4) – 5) + (13/4)(5 – 6) + (19/4)(6 – (23/4)]
Area of ∆ ADE = (1/2)[(4((23 – 20)/4) + (13/4)(– 1) + (19/4)((24 – 1)/4)]
Area of ∆ ADE = (1/2)[(4(3/4)) + (– 13/4) + (19/4)(1/4)]
Area of ∆ ADE = (1/2)[3 – 13/4 + 19/16]
Area of ∆ ADE = (1/2)((48 – 52 + 19)/16)
b) y1 = 6c) x2 = 13/4
d) y2 = 23/4
f) y3 = 5
Area of ∆ ADE = (1/2)[(4)((23/4) – 5) + (13/4)(5 – 6) + (19/4)(6 – (23/4)]
Area of ∆ ADE = (1/2)((67 – 52)/16)
Area of ∆ ADE = (1/2)((15)/16)
Area of ∆ ADE = 15/32 square units.------- equation 5
12) Now we will find the area of ∆ ABC with A(4, 6), B(1, 5), and E(7, 2).13) According to the problem,a) x1 = 4
b) y1 = 6c) x2 = 1
d) y2 = 5
e) x3 = 7
f) y3 = 2
14) We know that the area of the triangle ABC is as given below,
Area of ∆ ABC = (1/2)[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
Area of ∆ ABC = (1/2)[(4)(5 – 2) + (1)(2 – 6) + (7)(6 – 5)]
Area of ∆ ABC = (1/2)[(4(3) + (1)(– 4) + (7)(1)]
Area of ∆ ABC = (1/2)[(12)) + (– 4) + (7)]
Area of ∆ ABC = (1/2)[12 + 3]
Area of ∆ ABC = (1/2)(15)
b) y1 = 6c) x2 = 1
d) y2 = 5
f) y3 = 2
Area of ∆ ABC = (1/2)[(4)(5 – 2) + (1)(2 – 6) + (7)(6 – 5)]
Area of ∆ ABC = 15/2
Area of ∆ ABC = 15/2 square units.------- equation 6
15) Here we will find the ratio of the areas of these two triangles.
(Area of ∆ ADE)/(Area of ∆ ADE) = (15/32)/(15/2)
(Area of ∆ ADE)/(Area of ∆ ADE) = (15)(2)/(15)(32)
(Area of ∆ ADE)/(Area of ∆ ADE) = 2/32
(Area of ∆ ADE)/(Area of ∆ ADE) = 1/16
16) Hence, the ratio of the area of ∆ ADE to the area of ∆ ABC = 1:16.
Q 7. Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of ∆ ABC.
(i) The median from A meets BC at D. Find the coordinates of the point D.(ii) Find the coordinates of the point P on AD such that AP:PD = 2:1(iii) Find the coordinates of points Q and R on medians BE and CF
respectively such that BQ : QE = 2 : 1 and CR : RF = 2 : 1.
(iv) What do you observe?
[Note : The point which is common to all the three medians is called the centroid and this point divides each median in the ratio 2 : 1.]
(v) If A(x1, y1), B(x2, y2) and C(x3, y3) are the vertices of ∆ ABC, find the
coordinates of the centroid of the triangle.
Solution:
Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of ∆ ABC.
(i) The median from A meets BC at D. Find the coordinates of the point D.
1) The point D(x, y) the mid-point of the segment joining the points
B(6, 5), C(1, 4), so the coordinates of point D will be:
D(x, y) = ((6 + 1)/2, (5 + 4)/2)
D(x, y) = ((7)/2, (9)/2)
D(x, y) = (7/2, 9/2) -------- equation 1
2) The coordinates of point D are D(7/2, 9/2).
Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of ∆ ABC.
(ii) Find the coordinates of the point P on AD such that AP:PD = 2:1
1) The point P(x, y) divides the segment joining the points A(4, 2) and D(7/2, 9/2) in
the ratio 2 : 1, so,
we now, will find x coordinate
x = (2 (7/2) + 1 (4))/(2 + 1)x = (7 + 4)/(3)x = (11)/(3)x = 11/3 --------- equation 1we now, will find y coordinatey = (2 (9/2) + 1 (2))/(2 + 1)
y = (9 + 2)/(3)
y = (11)/(3)
y = 11/3 --------- equation 2
2) From equations 1 and 2, x = 11/3 and y = 11/3. So, the coordinates of P are P(11/3, 11/3).
Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of ∆ ABC.
(iii) Find the coordinates of points Q and R on medians BE and CF
respectively such that BQ : QE = 2 : 1 and CR : RF = 2 : 1.
1) First we will find the coordinates of midpoint E(x, y) of AC with A(4, 2), C(1, 4).
E(x, y) = ((4 + 1)/2, (2 + 4)/2)
E(x, y) = ((5)/2, (6)/2)
E(x, y) = (5/2, 3)
2) Now find the coordinates of point Q(x1, y1) which divides segment joining the
points B(6, 5) and E(5/2, 3) in the ratio 2:1.
we now, will find x coordinate
x = (2 (5/2) + 1 (6))/(2 + 1)
x = (5 + 6)/(3)
x = (11)/(3)
x = 11/3 --------- equation 1
we now, will find y coordinate
3) From equations 1 and 2, x = 11/3 and y = 11/3. So, the coordinates of
y = (2 (3) + 1 (5))/(2 + 1)
y = (6 + 5)/(3)
y = (11)/(3)
y = 11/3 --------- equation 2
Q are Q(11/3, 11/3).
4) Now we will find R which divides CF in 2:1.
5) First we will find the coordinates of midpoint F(x1, y1) of AB with A(4, 2), B(6, 5).
F(x1, y1) = ((4 + 6)/2, (2 + 5)/2)
F(x1, y1) = ((10)/2, (7)/2)
F(x1, y1) = (5, 7/2)
6) Now find the coordinates of point R(x, y) which divides segment joining the
points C(1, 4) and F(5, 7/2) in the ratio 2:1.
we now, will find x coordinate
x = (2 (5) + 1 (1))/(2 + 1)
x = (10 + 1)/(3)
x = (11)/(3)
x = 11/3 --------- equation 3
we now, will find y coordinate
7) From equations 3 and 4, x = 11/3 and y = 11/3. So, the coordinates of
y = (2 (7/2) + 1 (4))/(2 + 1)
y = (7 + 4)/(3)
y = (11)/(3)
y = 11/3 --------- equation 4
R are R(11/3, 11/3).
(iv) What do you observe?
[Note : The point which is common to all the three medians is called the centroid and this point divides each median in the ratio 2 : 1.]
1) From the above 2 sub-questions, we can say that points P, Q, and R are the
same. So we can say that all medians of the triangles are passing through one point.
(v) If A(x1, y1), B(x2, y2) and C(x3, y3) are the vertices of ∆ ABC, find the
coordinates of the centroid of the triangle.
1) Centroide of the ∆ ABC with A(x1, y1), B(x2, y2) and C(x3, y3) are:
P(x, y) = ((x1 + x2 + x3)/3, (y1 + y2 + y3)/3).
Q 8. ABCD is a rectangle formed by the points A(–1, –1), B(– 1, 4), C(5, 4) and
D(5, – 1). P, Q, R and S are the mid-points of AB, BC, CD, and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.
Solution:
1) P(x, y) is the midpoint of segment AB with A(-1, -1), B(-1, 4),
P(x, y) = ((-1 + (-1)/2, (-1 + 4)/2)
P(x, y) = ((-2)/2, (3)/2)
P(x, y) = (-1, 3/2)
2) Q(x, y) is the midpoint of segment BC with B(-1, 4), C(5, 4)
Q(x, y) = ((-1 + 5)/2, (4 + 4)/2)
Q(x, y) = ((4)/2, (8)/2)
Q(x, y) = (2, 4)
3) R(x, y) is the midpoint of segment CD with C(5, 4), D(5, -1),
R(x, y) = (5 + 5)/2, (4 -1)/2)
R(x, y) = (10/2, (3)/2)
R(x, y) = (5, 3/2)
4) S(x, y) is the midpoint of segment DA with D(5, -1), A(-1, -1)
S(x, y) = (5 - 1)/2, (-1 -1)/2)
S(x, y) = (4/2, (-2)/2)
S(x, y) = (2, -1)
5) Now we will find PQ with P(-1, 3/2), Q(2, 4)
a) x1 = -1
b) y1 = 3/2
c) x2 = 2
d) y2 = 4
6) Use the distance formula to get PQ:
(PQ) = √[(x1 – x2)2 + (y1 – y2)2]
(PQ) = √[(-1 - 2)2 + (3/2 - 4)2]
(PQ) = √[(- 3)2 + (- 5/2)2]
(PQ) = √[9 + (25/4)]
(PQ) = √[(36 + 25)/4]
(PQ) = √[(61)/4]
(PQ) = √61/2 ------------- equation 1
7) Now we will find QR with Q(2, 4), R(5, 3/2)
a) x1 = 2
b) y1 = 4
c) x2 = 5
d) y2 = 3/2
8) Use the distance formula to get QR:
(QR) = √[(x1 – x2)2 + (y1 – y2)2]
(QR) = √[(2 – 5)2 + (4 – 3/2)2]
(QR) = √[(– 3)2 + (5/2)2]
(QR) = √[9 + 25/4]
(QR) = √[(36 + 25)/4](QR) = √61/2 ------------- equation 2(QR) = √[(61)/4]
9) Now we will find RS with R(5, 3/2), S(2, -1)
a) x1 = 5
b) y1 = 3/2
c) x2 = 2
d) y2 = -1
10) Use the distance formula to get RS:
(RS) = √[(x1 – x2)2 + (y1 – y2)2]
(RS) = √[(5 – 2)2 + (3/2 – (-1)2]
(RS) = √[(3)2 + (3/2 + 1)2]
(RS) = √[(3)2 + (3 + 2)/2)2]
(RS) = √[(3)2 + (5)/2)2]
(RS) = √[9 + 25/4]
(RS) = √[(36 + 25)/4](RS) = √61/2 ------------- equation 3(RS) = √[(61)/4]
11) Now we will find SP with S(2, -1), P(-1, 3/2)
a) x1 = 2
b) y1 = -1
c) x2 = -1
d) y2 = 3/2
12) Use the distance formula to get SP:
(SP) = √[(x1 – x2)2 + (y1 – y2)2]
(SP) = √[(2 – (-1))2 + (-1 – 3/2)2]
(SP) = √[(3)2 + ((-2 -3)/2)2]
(SP) = √[(3)2 + (-5)/2)2]
(SP) = √[9 + 25/4]
(SP) = √[(36 + 25)/4](SP) = √61/2 ------------- equation 4(SP) = √[(61)/4]
13) From equations 1, 2, 3, and 4, we have,
PQ = QR = RS = SP ------------- equation 5
14) Now we will find PR with P(-1, 3/2), and R(5, 3/2)
a) x1 = -1
b) y1 = 3/2
c) x2 = 5
d) y2 = 3/2
15) We know that:
(PR) = √[(x1 – x2)2 + (y1 – y2)2]
(PR) = √[(-1 – 5)2 + (3/2 – 3/2)2]
(PR) = √[(– 6)2 + (0)2]
(PR) = √[36]
(PR) = 6 ------------- equation 6
16) Now we will find QS with Q(2, 4), and S(2, -1)
a) x1 = 2
b) y1 = 4
c) x2 = 2
d) y2 = -1
17) We know that:
(QS) = √[(x1 – x2)2 + (y1 – y2)2]
(QS) = √[(2 – 2)2 + (4 – (-1))2]
(QS) = √[(0)2 + (5)2]
(QS) = √[25]
(QS) = 5 ------------- equation 7
15) From equations 6 and 7, we have,
PR ≠ QS ------------- equation 8
16) From equations 5 and 8, we have,
all sides and diagonals are equal, so ABCD is the rhombus.
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