Chapter 7 Coordinate Geometry NCERT Exemplar Solutions Exercise 7.4 Class 10 Maths
Chapter Name  NCERT Maths Exemplar Solutions for Chapter 7 Coordinate Geometry Exercise 7.4 
Book Name  NCERT Exemplar for Class 10 Maths 
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Exercise 7.4 Solutions
Long Answer Questions
1. If (4, 3) and (4, 3) are two vertices of an equilateral triangle, find the coordinates of the third vertex, given that the origin lies in the interior of the triangle.
Solution
Let us take the vertices be (x, y)
Distance between (x, y) & (4, 3) is = √((x4)^{2} + (y3)^{2}) ...(1)
Distance between (x, y) & (4,3) is = √((x+4)^{2} + (y3)^{2}) ...(2)
Distance between (4,3) &(4,3) is =√((4+4)^{2} + (33)^{2})
= √(8)^{2}
= 8
As given in the question,
Equation (1) = (2)
(x4)^{2} = (x+4)^{2}
⇒ x^{2} 8x+16 = x^{2} +8x+16
⇒ 16x = 0
⇒ x = 0
Also, equation (1) = 8
(x4)^{2} +(y3)^{2} = 64 ...(3)
Putting the value of x in (3)
We get,
(04)^{2} +(y3)^{2} = 64
⇒ (y3)^{2} = 6416
⇒ (y3)^{2} = 48
⇒ y3 = √48
⇒ y = 3 + 4√3, 3  4√3
We do not take y = 3 + 4√3 as if y = 3 + 4√3 then origin cannot be at interior of triangle
Therefore, the third vertex = (0, 3  4√3)
2. A (6, 1), B (8, 2) and C (9, 4) are three vertices of a parallelogram ABCD. If E is the midpoint of DC, find the area of △ ADE.
Solution
As given in the question,
The three vertices of a parallelogram ABCD are A(6, 1), B(8, 2) and C(9, 4)
Let the fourth vertex of parallelogram = (x, y)
We know that, diagonals of a parallelogram bisect each other
Mid  point of a line segment joining the points (x_{1} , y_{1} ) and (x_{2} , y_{2} ) is given by,
So, fourth vertex is D(7, 3)
Now, Midpoint of DC,
(7 + 9)/2, (3 + 4)/2 = E
⇒ 8, 7/2 = E
Now,
Area of Î”ABC with vertices (x_{1} , y_{1} ) and (x_{2} , y_{2} ) and (x_{3} , y_{2} ) = ½ [ x_{1}(y_{2} – y_{3} ) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2}) ]
Area of Î”ADE with vertices A (6, 1), D(7, 3) and E(8, 7/2)
Putting value in the formula we get,
The required area of Î”ADE is 3/4 sq. units.
3. The points A (x_{1} , y_{1} ), B (x_{2} , y_{2} ) and C (x_{3} , y_{3} ) are 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 are the coordinates of the centroid of the triangle ABC?
Solution
Given,
The vertices of Î”ABC = A, B and C
Coordinates of A, B and C = A (x_{1} , y_{1} ), B (x_{2} , y_{2} ) , C (x_{3} , y_{3} )
(i) As given in the question D is the mid  point of BC and it bisect the line into two equal parts.
Coordinates of the mid  point of BC;
D = [(x_{2} + x_{3} )/2 , (y_{2} + y_{3} )/2]
(ii) Let the coordinates of a point P be (x, y)
Given,
The ratio in which the point P(x, y), divide the line joining,
A (x_{1} , y_{1}) and D = [(x_{2} + x_{3})/2 , (y_{2} + y_{3})/2] = 2 : 1
Then,
Coordinates of P =
(iii) Let the coordinates of a point Q be (p, q)
Given,
The point Q (p, q), divides the line B and E in 2 : 1,
Also,
Midpoint of AC = Coordinates of E (As BE is median of CE),
E = [ (x_{1} , x_{3} )/2 , (y_{1} , y_{3} )/2]
Also point R divides the line CF in the ratio 2 : 1,
So coordinates of R =
R = [(Sum of all coordinates of all vertices)/3, (Sum of all coordinates of all vertices)/3]
R = [(x_{2} + x_{3} + x_{1} )/3, (y_{2} + y_{3} + y_{1})/3]
In parallelogram, we know that, diagonals are bisects each other so,
Midpoint of AC = midpoint of BD
[(1 + a)/2 , (2+2)/2] = [(2  4)/2, (3  3)/2]
⇒ a = 1
So, the required value of a is – 3.
Taking slope of DP m_{2} , as DP is perpendicular to AB, we can apply the condition,
m_{1} . m_{2} = 1
⇒ 5.m_{2} = 1
⇒ m_{2} = 1/5
Now, equation of DP having slope m_{2} , passing through point (4, 3) is,
y  y_{1} = m_{2} (x  x_{1} )
⇒ y + 3 = (1/5)(x + 4)
⇒ x + 5y = 19
Now, on solving both the equations we ge,
x = 8/13
y = 51/13
Coordinates of P = (8/13, 51/13)
So, height of parallelogram,
BD = 8
As,
AB = BC = CD = DA
Also, AC = BD
We see that, AB = BC = CD = DA, all sides are equal.
[(3 + 11)/2, (5+5)/2]
(7, 5)
In the figure we had located every place with its coordinates and direction,
By distance formula,
= √[(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}]
Distance between house and the bank = √[(5 – 2)^{2} + (8 – 4)^{2}] = 5
Distance between bank and the daughter school = √[(13 – 5)^{2} + (14 – 8)^{2}] = 10
Distance between the daughter school and office = √[(13 – 13)^{2}  (26 – 14)^{2}] = 12
Now,
Total distance = 5 + 10 + 12 = 27 units
Distance between house to offices = √[(13 – 2)^{2} + (26 – 4)^{2}] = 24.6 km