﻿ Class 10 NCERT Math Solution
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TOPICS
Exercise - 7.4 (Optional)

Question-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).

Solution :-
```  Let the given line divide the line segment joining the points A(2, −2) and B(3, 7) in a ratio k:1.
Coordinates of the point of division = [(3k+2)/(k+1), (7k-2)/(k+1)]
This point also lies on 2x + y − 4 = 0
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
Therefore, the ratio in which the line 2x + y − 4 = 0 divides the line segment joining the points A(2, −2) and B(3, 7) is 2:9.
```

Question-2 :-  Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.

Solution :-
```  If the given points are collinear, then the area of triangle formed by these points will be 0.
By using the area of triangle formula, is given by
0 = 1/2[x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)]
0 = 1/2[x(2 - 0) + 1(0 - y) + 7(y - 2)]
0 = 1/2[2x - y + 7y - 14]
0 = 1/2[2x + 6y - 14]
0 = 1/2 x 2[x + 3y - 7]
0 = x + 3y - 7
x + 3y - 7 = 0
This is the required relation between x and y.
```

Question-3 :-  Find the centre of a circle passing through the points (6, – 6), (3, – 7) and (3, 3).

Solution :-
```  Let O (x, y) be the centre of the circle.
And let the points (6, −6), (3, −7), and (3, 3) be representing the points A, B, and C on the circumference of the circle.
Therefore,
OA = √(x-6)² + (y+6)²
OB = √(x-3)² + (y+7)²
OC = √(x-3)² + (y-3)²

However, OA = OB (Radii of the same circle)
√(x-6)² + (y+6)² = √(x-3)² + (y+7)²
(x-6)² + (y+6)² = (x-3)² + (y+7)²
x² + 36 - 12x + y² + 36 + 12y = x² + 9 - 6x + y² + 49 + 14y
72 - 12x + 12y = 58 - 6x + 14y
-12x + 6x + 12y - 14y = 58 - 72
-6x - 2y = -14
3x + y = 7  .....(i)

Similarly, OA = OC (Radii of the same circle)
√(x-6)² + (y+6)² = √(x-3)² + (y-3)²
(x-6)² + (y+6)² = (x-3)² + (y-3)²
x² + 36 - 12x + y² + 36 + 12y = x² + 9 - 6x + y² + 9 - 6y
72 - 12x + 12y = 18 - 6x - 6y
-12x + 6x + 12y + 6y = 18 - 72
-6x + 18y = -54
-3x + 9y = -27 .....(ii)

By adding eq. (i) and eq. (ii),
3x + y - 3x + 9y = 7 - 27
10y = -20
y = -2
Put in eq. (i),
3x - 2 = 7
3x = 7 + 2
x = 9/3
x = 3

Therefore, the centre of the circle is (3, −2).
```

Question-4 :-  The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices.

Solution :-
```  Let ABCD be a square having (−1, 2) and (3, 2) as vertices A and C respectively.
Let (x, y), (x₁, y₁) be the coordinate of vertex B and D respectively. We know that the sides of a square are equal to each other.
Therefore, AB = BC
√(x+1)² + (y-2)² = √(x-3)² + (y-2)²
x² + 2x + 1 + y² - 4y + 4 = x² + 9 - 6x + y² + 4 - 4y
5 + 2x = 13 - 6x
2x + 6x = 13 - 5
8x = 8
x = 1

We know that in a square, all interior angles are of 90°.
Now, In ΔABC,
AB² + BC² = AC²
[√(1+1)² + (y-2)²]² + [√(1-3)² + (y-2)²]² = [√(3+1)² + (2-2)²]²
4 + y² - 4y + 4 + 4 + y² - 4y + 4 = 16
16 + 2y² - 8y = 16
2y² - 8y = 0
2y(y - 4) = 0
2y = 0; y - 4 = 0
y = 0; y = 4

We know that in a square, the diagonals are of equal length and bisect each other at 90°.
Let O be the mid-point of AC.
Therefore, it will also be the mid-point of BD.
Coordinate of point O = [(-1+3)/2, (2+2)/2] = (1, 2)
Now, [(1+x₁)/2, (y+y₁)/2] = (1, 2)
(1+x₁)/2 = 1
1 + x₁ = 2
x₁ = 2 - 1
x₁ = 1

(y+y₁)/2 = 2
y + y₁ = 4
if, y = 0, then y₁ = 4
if, y = 4, then y₁ = 0
Therefore, the required coordinates are (1, 0) and (1, 4).
```

Question-5 :-  The Class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Sapling 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 Figure. 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 Δ PQR if C is the origin? Also calculate the areas of the triangles in these cases. What do you observe? Solution :-
```(i) Taking A as origin, we will take AD as x-axis and AB as y-axis.
It can be observed that the coordinates of point P, Q, and R are (4, 6), (3, 2), and (6, 5) respectively.

Now, Area of triangle PQR
= 1/2[x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)]
= 1/2[4(2 - 5) + 3(5 - 6) + 6(6 - 2)]
= 1/2[-12 - 3 + 24]
= 1/2 x 9
= 9/2
Therefore, area of triangle PQR = 9/2 square units.
```
```(ii) Taking C as origin, CB as x-axis, and CD as y-axis.
The coordinates of vertices P, Q, and R are (12, 2), (13, 6), and (10, 3) respectively.

Now, Area of triangle PQR
= 1/2[x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)]
= 1/2[12(6 - 3) + 13(3 - 2) + 10(2 - 6)]
= 1/2[36 + 13 - 40]
= 1/2 x 9
= 9/2
Therefore, area of triangle PQR = 9/2 square units.

It can be observed that the area of the triangle is same in both the cases.
```

Question-6 :-  The vertices of a Δ 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 Δ ADE and compare it with the area of Δ ABC.

Solution :-
```  Given that : AD/AB = AE/AC = 1/4 Therefore, D and E are two points on side AB and AC respectively such that they divide side AB and AC in a ratio of 1:3.

Coordinates of point D = [(1x1 + 3x4)/(1+3), (1x5 + 3x6)/(1+3)] = [(1+12)/4, (5+18)/4] = (13/4, 23/4)
Coordinates of point E = [(1x7 + 3x4)/(1+3), (1x2 + 3x6)/(1+3)] = [(7+12)/4, (2+18)/4] = (19/4, 20/4)

= 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

Now, Area of triangle ABC
= 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]
= 1/2 x 15
= 15/2 square units

Clearly, the ratio between the areas of ΔADE and ΔABC is 1:16.
```

Question-7 :-  Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of Δ 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 yo 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 Δ ABC, find the coordinates of the centroid of the triangle.

Solution :-
``` (i) Median AD of the triangle will divide the side BC in two equal parts.
Therefore, D is the mid-point of side BC.
Coordinates of D = [(6+1)/2, (5+4)/2] = (7/2, 9/2)
```
```(ii) Point P divides the side AD in a ratio 2:1.
Coordinates of P = [(2x7/2 + 1x4)/(2+1), (2x9/2 + 1x2)/(2+1)] = (11/3, 11/3)
```
```(iii) Median BE of the triangle will divide the side AC in two equal parts.
Therefore, E is the mid-point of side AC.
Coordinates of E = [(4+1)/2, (2+4)/2] = (5/2, 3)

Point Q divides the side BE in a ratio 2:1.
Coordinates of Q = [(2x5/2 + 1x6)/(2+1), (2x3 + 1x5)/(2+1)] = (11/3, 11/3)

Median CF of the triangle will divide the side AB in two equal parts.
Therefore, F is the mid-point of side AB.
Coordinates of F = [(4+6)/2, (2+5)/2] = (5, 7/2)

Point R divides the side CF in a ratio 2:1.
Coordinates of R = [(2x5 + 1x1)/(2+1), (2x7/2 + 1x4)/(2+1)] = (11/3, 11/3)
```
```(iv) It can be observed that the coordinates of point P, Q, R are the same.
Therefore, all these are representing the same point on the plane.
i.e., the centroid of the triangle.
```
```(v) Consider a triangle, ΔABC, having its vertices as A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃).
Median AD of the triangle will divide the side BC in two equal parts.
Therefore, D is the mid-point of side BC.

Coordinates of D = [(x₂+x₃)/2, (y₂+y₃)/2]
Let the centroid of this triangle be O.
Point O divides the side AD in a ratio 2:1.
Coordinates of O = [(x₁+x₂+x₃)/2, (y₁+y₂+y₃)/2]
```

Question-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 :-
```  P is the midpoint of side AB. Therefore, the coordinates of P are [(-1-1)/2, (-1+4)/2] = (-1, 3/2)
Similarly, the coordinates of Q, R and S are (2, 4), (5, 3/2) and (2, -1) respectively.
Length of PQ = √(-1-2)² + (3/2-4)² = √61/4
Length of QR = √(2-5)² + (4-3/2)² = √61/4
Length of RS = √(5-2)² + (3/2+1)² = √61/4
Length of SP = √(2+1)² + (-1-3/2)² = √61/4
Length of PR = √(-1-5)² + (3/2-3/2)² = 6
Length of QS = √(2-2)² + (4+1)² = 5
It can be observed that all sides of the given quadrilateral are of the same measure.
However, the diagonals are of different lengths. Therefore, PQRS is a rhombus.
```
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