Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry Notes PDF Download: Tamil Nadu STD 10th Maths Chapter 5 Coordinate Geometry Notes 
Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry Notes PDF Download: Students of class can download the Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry Notes PDF Download from our website. We have uploaded the Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry notes according to the latest chapters present in the syllabus. Download Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry Chapter Wise Notes PDF from the links provided in this article.
Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry Notes PDF Download
We bring to you specially curated Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry Notes PDF which have been prepared by our subject experts after carefully following the trend of the exam in the last few years. The notes will not only serve for revision purposes, but also will have several cuts and easy methods to go about a difficult problem.
Board 
Tamilnadu Board 
Study Material 
Notes 
Class 
Samacheer Kalvi 10th Maths 
Subject 
10th Maths 
Chapter 
Chapter 5 Coordinate Geometry 
Format 

Provider 
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Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 5 Coordinate Geometry Unit Exercise 5
Question 1.
PQRS is a rectangle formed by joining the points P(1, 1), Q(1, 4) ,R(5, 4) and S(5, 1). A, B, C and D are the midpoints of PQ, QR, RS and SP respectively. Is the quadrilateral ABCD a square, a rectangle or a rhombus? Justify your answer.
Solution:
A, B, C and D are mid points of PQ, QR, RS & SP respectively.
∴ AB and BC are not perpendicular
⇒ ABCD is rhombus as diagonals are perpendicular and sides are not perpendicular.
Question 2.
The area of a triangle is 5 sq.units. Two of its vertices are (2, 1) and (3, 2). The third Vertex is (x, y) where y = x + 3 . Find the coordinates of the third vertex.
Solution:
Area of triangle formed by points (x_{1}, y_{1}),
Question 3.
Find the area of a triangle formed by the lines 3x + y – 2 = 0, 5x + 2y – 3 = 0 and 2x – y – 3 = 0
Solution:
Question 4.
If vertices of a quadrilateral are at A(5, 7), B(4, k) , C(1, 6) and D(4, 5) and its area is
72 sq.units. Find the value of k.
Area (quadrilateral ABCD)
Question 5.
Without using distance formula, show that the points (2, 1) , (4, 0) , (3, 3) and (3, 2) are vertices of a parallelogram.
Solution:
Slope of AB = Slope of CD
Slope of BC = Slope of DA
Hence ABCD forms a parallelogram.
Question 6.
Find the equations of the lines, whose sum and product of intercepts are 1 and 6 respectively.
Let the intercepts be x_{1}, y_{1} respectively
Question 7.
The owner of a milk store finds that, he can sell 980 litres of milk each week at ₹ 14/litre and 1220 litres of milk each week at ₹ 16 litre. Assuming a linear relationship
between selling price and demand, how many litres could he sell weekly at ₹ 17/litre?
Solution:
Question 8.
Find the image of the point (3, 8) with respect to the line x + 3y = 7 assuming the line to be a plane mirror.
Solution:
Question 9.
Find the equation of a line passing through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes.
Solution:
4x + 7y – 3 = 0
2x – 3y + 1 = 0
4x + 7y – 3 – 2(2x – 3y + 1) = 0
Question 10.
A person standing at a junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 seek to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find the equation of the path that he should follow.
Solution:
Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 5 Coordinate Geometry Additional Questions
Question 1.
Find a relation between x and y such that the point (x, y) is equidistant from the points (7, 1) and (3, 5).
Solution:
Let P(x, y) be equidistant from the points A(7, 1) and B(3, 5).
We are given that AP = BP. So, AP^{2} = BP^{2}
(x – 7)^{2} + (y – 1)^{2} =(x – 3)^{2} + (y – 5)^{2}
x^{2} – 14x + 49 + y^{2} – 2y + 1 = x^{2} – 6x + 9 + y^{2} – 10y + 25
x – y = 2
Which is the required relation
Question 2.
Show that the points (1, 7), (4, 2), (1, 1) and (4, 4) are the vertices of a square.
Solution:
Let A(1, 7), B(4, 2), C(1, 1) and D(4, 4) be the given points. To prove that ABCD is a square, we have to prove that all its sides are equal and both its diagonals are equal.
Since, AB = BC = CD = DA and AC = BD, all the four sides of the quadrilateral ABCD are equal and its diagonals AC and BD are also equal. Therefore, ABCD is a square.
Question 3.
If A (5, 7), B (4, 5), C (1, 6) and D (4, 5) are the vertices of a quadrilateral, find the area of the quadrilateral ABCD.
Solution:
By joining B to D, you will get two triangles ABD and BCD.
Now, the area of ∆ABD
So, the area of quadrilateral ABCD = 53 + 19 = 72 square units.
Question 4.
Find the coordinates of the points of trisection (i.e. points dividing in three equal parts) of the line segment joining the points A(2, 2) and B(7, 4).
Solution:
Let P and Q be the points of trisection at AB. i.e., AP = PQ = QB
Therefore, P divides AB internally in the ratio 1 : 2. Therefore, the coordinates at P, by applying the section formula, are
Now, Q also divides AB internally in the ratio 2 : 1, so, the coordinates at Q are
Therefore, the coordinates of the points at trisection of the line segment joining A and B are (1, 0) and (4, 2).
Question 5.
If the points A(6, 1), B(8, 2), C(9, 4) and D(P, 3) are the vertices of a parallelogram, taken in order. Find the value of P.
Solution:
We know that diagonals of a parallelogram bisect each other.
So, the coordinates at the midpoint of AC = coordinates of the midpoint of BD.
Question 6.
Find the area of a triangle whose vertices are (1,1), (4, 6) and (3, 5).
Solution:
The area of the triangle formed by the vertices A(1, 1), B(4, 6) and C(3, 5), by using the formula
So, the area of the triangle is 24 square units.
Question 7.
If A(2, 1), B(a, 0), C(4, b) and D(1, 2) are the vertices of a parallelogram, find the values of a and b.
Solution:
We know that the diagonals of a parallelogram bisect each other. Therefore the coordinates of the midpoint of AC are same as the coordinates of the midpoint of BD. i.e.
Question 8.
Find the area of the quadrilateral whose vertices, taken in order, are (3, 2), (5, 4), (7, 6) and (5, 4).
Solution:
We have Area of the quadrilateral
Question 9.
Find the area of the triangle formed by the points P(1.5, 3), Q(6, 2) and R(3, 4).
Solution:
The area of the triangle formed by the given points is equal to
Can we have a triangle of area 0 square units? What does this mean?
If the area of a triangle is 0 square units, then its vertices will be collinear.
Question 10.
Find the value of k if the pointsA(2, 3), B(4, k) and (6, 3) are collinear.
Since the given points are collinear, the area a the triangle formed by them must be 0, i.e.
How to Prepare using Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry Notes PDF?
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 Use Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry notes by paying attention to facts and ideas.
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