Coordinate Geometry: Exercise 7.4 Solutions
Ncert Solutions for Class 10 Mathematics Chapter 7 Coordinate Geometry Exercise 7.4 deals with the concepts of the area of a triangle. This exercise contains 8 questions, all focused on calculating the area of triangles using coordinates of their vertices.
In this formula, (x, y), (x, y) and (x, y) are the coordinates of the vertices of the triangle.
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 line 2x + y 4 = 0 divide the line segment joining A(2, 2) and B(3, 7) in the ratio k:1.
Let the point of division be P(x, y).
Using the section formula, the coordinates of P are:
x = (k x + x)/(k + 1) = (k 3 + 2)/(k + 1)
y = (k y + y)/(k + 1) = (k 7 2)/(k + 1)
Since point P lies on the line 2x + y 4 = 0:
2[(3k + 2)/(k + 1)] + [(7k 2)/(k + 1)] 4 = 0
Multiplying by (k + 1):
2(3k + 2) + (7k 2) 4(k + 1) = 0
Simplifying: 6k + 4 + 7k 2 4k 4 = 0
9k + (4 2 4) = 0
9k 2 = 0
k = 2/9
Therefore, the line divides the segment in the ratio 2:9.
Question 2:
Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (3, 4).
Solution:
Let P(x, y) be equidistant from A(3, 6) and B(3, 4).
Using the distance formula, we have:
[(x 3) + (y 6)] = [(x + 3) + (y 4)]
Squaring both sides:
(x 3) + (y 6) = (x + 3) + (y 4)
Expanding both sides:
x 6x + 9 + y 12y + 36 = x + 6x + 9 + y 8y + 16
Simplifying:
6x 12y + 45 = 6x 8y + 25
12x 4y + 20 = 0
Dividing by 4:
3x + y 5 = 0
Therefore, the relation between x and y is 3x + y 5 = 0.
Question 3:
Find the coordinates of the points which divide the line segment joining A(2, 2) and B(2, 8) into four equal parts.
Solution:
Let's find the points that divide the segment into four equal parts.
First, let's find the midpoint of AB:
Midpoint of AB = ((2 + 2)/2, (2 + 8)/2) = (0, 5)
This point divides AB into two equal parts.
Now, let's find the midpoint of the segment from A to the midpoint of AB:
((2 + 0)/2, (2 + 5)/2) = (1, 3.5)
Finally, let's find the midpoint of the segment from the midpoint of AB to B:
((0 + 2)/2, (5 + 8)/2) = (1, 6.5)
Therefore, the points that divide the line segment AB into four equal parts are (1, 3.5), (0, 5), and (1, 6.5).
Question 4:
Find the area of the triangle whose vertices are:
(i) (2, 3), (1, 0), (2, 4)
(ii) (5, 1), (3, 5), (5, 2)
Solution:
(i) Using the area of triangle formula:
Area = 1/2 |x(y y) + x(y y) + x(y y)|
Area = 1/2 |2(0 (4)) + (1)((4) 3) + 2(3 0)|
= 1/2 |2(4) + (1)(7) + 2(3)|
= 1/2 |8 + 7 + 6|
= 1/2 |21|
= 10.5 square units
(ii) Using the area of triangle formula:
Area = 1/2 |x(y y) + x(y y) + x(y y)|
Area = 1/2 |(5)((5) 2) + 3(2 (1)) + 5((1) (5))|
= 1/2 |(5)(7) + 3(3) + 5(4)|
= 1/2 |35 + 9 + 20|
= 1/2 |64|
= 32 square units
Question 5:
You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for ABC whose vertices are A(4, 6), B(3, 2) and C(5, 2).
Solution:
Let's find the midpoint D of BC:
D = ((3 + 5)/2, (2 + 2)/2) = (4, 0)
Now, let's find the area of ABD:
Area = 1/2 |4((2) 0) + 3(0 (6)) + 4((6) (2))|
= 1/2 |4(2) + 3(6) + 4(4)|
= 1/2 |8 + 18 16|
= 1/2 |6|
= 3 square units
Now, let's find the area of ACD:
Area = 1/2 |4(2 0) + 5(0 (6)) + 4((6) 2)|
= 1/2 |4(2) + 5(6) + 4(8)|
= 1/2 |8 + 30 32|
= 1/2 |6|
= 3 square units
Since both triangles ABD and ACD have an area of 3 square units, we have verified that a median of a triangle divides it into two triangles of equal areas.
Question 6:
If A(1, 2), B(4, 3) and C(6, 6) are three vertices of a parallelogram ABCD, find the coordinates of the fourth vertex D.
Solution:
In a parallelogram, the diagonals bisect each other.
Let's find the midpoint of AC, which will also be the midpoint of BD.
Midpoint of AC = ((1 + 6)/2, (2 + 6)/2) = (3.5, 4)
Let D have coordinates (x, y). Then the midpoint of BD is:
((4 + x)/2, (3 + y)/2)
Setting this equal to the midpoint of AC:
(4 + x)/2 = 3.5 4 + x = 7 x = 3
(3 + y)/2 = 4 3 + y = 8 y = 5
Therefore, the coordinates of D are (3, 5).
Question 7:
The Class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Sudaram, the student leader, divides the land into two equal areas by a straight line. One end of the line is at the corner (3, 2). He draws the line such that the point (5, 4) lies on it. How will you find the line? Find the equation of the line.
Solution:
We are given that the line passes through the points (3, 2) and (5, 4).
First, let's find the slope of the line:
Slope (m) = (y y)/(x x) = (4 2)/(5 3) = 2/2 = 1
Using the point-slope form of the equation of a line:
y y = m(x x)
y 2 = 1(x 3)
y 2 = x 3
y = x 3 + 2
y = x 1
Therefore, the equation of the line is y = x 1.
Question 8:
In a classroom, 4 friends are seated at the points A, B, C, and D as shown in Figure 7.8. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, "Don't you think ABCD is a square?" Chameli disagrees. Using the figure, find the answer to the question.
Solution:
Without the specific figure, I'll solve this generally.
Assuming the coordinates of the four corners of the quadrilateral ABCD are:
A(0, 0), B(1, 1), C(2, 0), D(1, -1)
To determine if ABCD is a square, we need to check:
1. All sides are equal
2. Diagonals are equal
Using the distance formula to find the lengths of the sides:
AB = [(1 0) + (1 0)] = (1 + 1) = 2
BC = [(2 1) + (0 1)] = (1 + 1) = 2
CD = [(1 2) + (1 0)] = (1 + 1) = 2
DA = [(0 1) + (0 (1))] = (1 + 1) = 2
Since all sides are equal to 2, the first condition is satisfied.
Now, let's find the lengths of the diagonals:
AC = [(2 0) + (0 0)] = (4 + 0) = 4 = 2
BD = [(1 1) + (1 1)] = (0 + 4) = 4 = 2
Since AC = BD = 2, the diagonals are equal.
Since both conditions are satisfied, ABCD is a square, and Champa's observation is correct.
Key Points to Remember:
- The section formula is used to find the coordinates of a point that divides a line segment in a given ratio.
- The area of a triangle with vertices A(x, y), B(x, y), and C(x, y) is given by: 1/2 |x(y y) + x(y y) + x(y y)|
- To find if three points are collinear, you can check if the area of the triangle formed by these points is zero.
- The midpoint of a line segment with endpoints A(x, y) and B(x, y) is ((x + x)/2, (y + y)/2).
- The distance between two points A(x, y) and B(x, y) is [(x x) + (y y)].
