NCERT Solutions Class 10th Maths Chapter – 7 Coordinate Geometry Exercise – 7.4

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: Consider line 2x + y – 4 = 0 divides line AB joined by the two points A(2, -2) and B(3, 7) in k : 1 ratio.
Coordinates of point of division can be given as follows:
x = (2 + 3k)/(k + 1) and y = (-2 + 7k)/(k + 1)
Substituting the values of x and y given equation, i.e. 2x + y – 4 = 0, we have
2{(2 + 3k)/(k + 1)} + {(-2 + 7k)/(k + 1)} – 4 = 0
(4 + 6k)/(k + 1) + (-2 + 7k)/(k + 1) = 4
4 + 6k – 2 + 7k = 4(k+1)
-2 + 9k = 0
Or k = 2/9
Hence, the ratio is 2: 9.

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

Solution: If given points are collinear then area of triangle formed by them must be zero.
Let (x, y), (1, 2) and (7, 0) are vertices of a triangle,
Area of a triangle = 1/2 × [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)] = 0
[x(2 – 0) + 1 (0 – y) + 7( y – 2)] = 0
2x – y + 7y – 14 = 0
2x + 6y – 14 = 0
x + 3y – 7 = 0.
Which is the required result.

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

Solution: Let A = (6, -6), B = (3, -7), C = (3, 3) are the points on a circle.
If O is the centre, then OA = OB = OC (radii are equal)
If O = (x, y) then
OA = √[(x – 6)² + (y + 6)²]
OB = √[(x – 3)2 + (y + 7)²]
OC = √[(x – 3)2 + (y – 3)²]
Choose: OA = OB, we have
After simplifying above, we get -6x = 2y – 14 ….(1)
Similarly: OB = OC
(x – 3)² + (y + 7)² = (x – 3)²+ (y – 3)²
(y + 7)²2 = (y – 3)²
y²+ 14y + 49 = y²– 6y + 9
20y =-40
or y = -2
Substituting the value of y in equation (1), we get;
-6x = 2y – 14
-6x = -4 – 14 = -18
x = 3
Hence, the centre of the circle located at point (3,-2).

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 is a square, where A(-1,2) and B(3,2). And Point O is the point of intersection of AC and BD
To Find: Coordinate of points B and D.

Step 1: Find distance between A and C and coordinates of point O.
We know that, diagonals of a square are equal and bisect each other.
AC = √[(3 + 1)² + (2 – 2)²] = 4
Coordinates of O can be calculated as follows:
x = (3 – 1)/2 = 1 and y = (2 + 2)/2 = 2
So, O(1,2)
Step 2: Find the side of the square using Pythagoras theorem
Let a be the side of square and AC = 4
From right triangle, ACD,
a = 2√2
Hence, each side of square = 2√2
Step 3: Find coordinates of point D
Equate length measure of AD and CD
Say, if coordinate of D are (x₁, y₁)
AD =  √[(x₁ + 1)² + (y₁ – 2)²]
Squaring both sides,
AD² = (x1 + 1)²+ (y1 – 2)²
Similarly,  CD² = (x₁ – 3)² + (y₁ – 2)²
Since all sides of a square are equal, which means AD = CD
(x₁ + 1)² + (y₁ – 2)² = (x₁ – 3)² + (y₁ – 2)²
x₁² + 1 + 2x₁ = x₁² + 9 – 6x₁
8x₁ = 8
x₁ = 1
Value of y1 can be calculated as follows by using the value of x.
From step 2: each side of square = 2√2
CD² = (x₁ – 3)² + (y₁ – 2)²
8 = (1 – 3)² + (y₁ – 2)²
8 = 4 + (y₁ – 2)²
y₁ – 2 = 2
y₁= 4
Hence, D = (1, 4)
Step 4: Find coordinates of point B
From line segment, BOD
Coordinates of B can be calculated using coordinates of O; as follows:
Earlier, we had calculated O = (1, 2)
Say B = (x₂, y₂)
For BD;
1 = (x₂ + 1)/2
x₂ = 1
And 2 = (y₂ + 4)/2
=> y₂ = 0
Therefore, the coordinates of required points are B = (1,0) and D = (1,4)

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 1 m from each other. There is a triangular lawn in the plot as shown in the fig. 7.14. The students are to sow the 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.

Solution: Taking A as origin, coordinates of the vertices P, Q and R are,
From figure: P = (4, 6), Q = (3, 2), R (6, 5)
Here AD is the x-axis and AB is the y-axis.

(ii) What will be the coordinates of the vertices of triangle PQR if C is the origin?

Solution: Taking C as origin,
Coordinates of vertices P, Q and R are ( 12, 2), (13, 6) and (10, 3) respectively.
Here CB is the x-axis and CD is the y-axis.
Find the area of triangles:
Area of triangle PQR in case of origin A:
Using formula: Area of a triangle = 1/2 × [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= ½ [4(2 – 5) + 3 (5 – 6) + 6 (6 – 2)]
= ½ (- 12 – 3 + 24 )
= 9/2 sq unit

(iii) Also calculate the areas of the triangles in these cases. What do you observe?

Solution: Area of triangle PQR in case of origin C:
Area of a triangle = 1/2 × [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= ½ [ 12(6 – 3) + 13 ( 3 – 2) + 10( 2 – 6)]
= ½ ( 36 + 13 – 40)
= 9/2 sq unit
This implies, Area of triangle PQR at origin A = Area of triangle PQR at origin C
Area is same in both case because triangle remains the same no matter which point is considered as origin.

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 area of ∆ ABC. (Recall Theorem 6.2 and Theorem 6.6)

Solution: Given: The vertices of a ∆ ABC are A (4, 6), B (1, 5) and C (7, 2)

AD/AB = AE/AC = 1/4
AD/(AD + BD) = AE/(AE + EC) = 1/4
Point D and Point E divide AB and AC respectively in ratio 1 : 3.
Coordinates of D can be calculated as follows:
x = (m₁x₂ + m₂x₁)/(m₁ + m₂) and y = (m₁y₂ + m₂y₁)/(m₁ + m₂)
Here m1 = 1 and m2 = 3
Consider line segment AB which is divided by the point D at the ratio 1:3.
x = [3(4) + 1(1)]/4 = 13/4
y = [3(6) + 1(5)]/4 = 23/4
Similarly, Coordinates of E can be calculated as follows:
x = [1(7) + 3(4)]/4 = 19/4
y = [1(2) + 3(6)]/4 = 20/4 = 5
Find Area of triangle:
Using formula: Area of a triangle  = 1/2 × [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
Area of triangle ∆ ABC can be calculated as follows:
= ½ [4(5 – 2) + 1( 2 – 6) + 7( 6 – 5)]
= ½ (12 – 4 + 7) = 15/2 sq unit
Area of ∆ ADE can be calculated as follows:
= ½ [4(23/4 – 5) + 13/4 (5 – 6) + 19/4 (6 – 23/4)]
= ½ (3 – 13/4 + 19/16)
= ½ ( 15/16 ) = 15/32 sq unit
Hence, ratio of area of triangle ADE to area of triangle ABC = 1 : 16.

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 point D.

Solution: Coordinates of D can be calculated as follows:
Coordinates of D = ( (6+1)/2, (5+4)/2 ) = (7/2, 9/2)
So, D is (7/2, 9/2)

(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1.

Solution: Coordinates of P can be calculated as follows:
Coordinates of P = ( [2(7/2) + 1(4)]/(2 + 1), [2(9/2) + 1(2)]/(2 + 1) ) = (11/3, 11/3)
So, P is (11/3, 11/3)

(iii) Find the coordinates of point Q and R on medians BE and CF respectively such that BQ : QE = 2:1 and CR : RF = 2 : 1.

Solution: Coordinates of E can be calculated as follows:
Coordinates of E = ( (4+1)/2, (2+4)/2 ) = (5/2, 6/2) = (5/2, 3)
So, E is (5/2, 3)
Point Q and P would be coincident because medians of a triangle intersect each other at a common point called centroid. Coordinate of Q can be given as follows:
Coordinates of Q =( [2(5/2) + 1(6)]/(2 + 1), [2(3) + 1(5)]/(2 + 1) ) = (11/3, 11/3)
F is the mid- point of the side AB
Coordinates of F = ( (4+6)/2, (2+5)/2 ) = (5, 7/2)
Point R divides the side CF in ratio 2:1
Coordinates of R = ( [2(5) + 1(1)]/(2 + 1), [2(7/2) + 1(4)]/(2 + 1) ) = (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.]

Solution: Coordinates of P, Q and R are same which shows that medians intersect each other at a common point, i.e. centroid of the triangle.

(v) If A (x₁, y₁), B (x₂, y₂) and C (x₃, y₃) are the vertices of triangle ABC, find the coordinates of the centroid of the triangle.

Solution: If A (x₁, y₁), B (x₂, y₂) and C (x₃, y₃) are the vertices of triangle ABC, the coordinates of centroid can be given as follows:
x = (x₁ + x₂ + x₃)/3 and y = (y₁ + y₂ + y₃)/3

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 midpoints of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.

Solution:

P id the mid-point of side AB,
Coordinate of P = ( (-1 – 1)/2, (-1 + 4)/2 ) = (-1, 3/2)
Similarly, Q, R and S are (As Q is mid-point of BC, R is midpoint of CD and S is midpoint of AD)
Coordinate of Q = (2, 4)
Coordinate of R = (5, 3/2)
Coordinate of S = (2, -1)
Now,
Length of PQ = √[(-1 – 2)² + (3/2 – 4)²] = √(61/4) = √61/2
Length of SP = √[(2 + 1)² + (-1 – 3/2)²] = √(61/4) = √61/2
Length of QR = √[(2 – 5)² + (4 – 3/2)²] = √(61/4) = √61/2
Length of RS = √[(5 – 2)² + (3/2 + 1)²] = √(61/4) = √61/2
Length of PR (diagonal) = √[(-1 – 5)² + (3/2 – 3/2)²] = 6
Length of QS (diagonal) = √[(2 – 2)² + (4 + 1)²] = 5
The above values show that, PQ = SP = QR = RS = √61/2, i.e. all sides are equal.
But PR ≠ QS i.e. diagonals are not of equal measure.
Hence, the given figure is a rhombus.