In This Post we are providing Chapter 7Coordinate Geometry NCERT Solutions for Class 10 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These Coordinate Geometry Class 10 solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.
We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 10 Maths Coordinate Geometry NCERT Written Solutions & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.
Exercise 7.1
1. Find the distance between the following pairs of points:
(i) (2, 3), (4, 1) (ii) (−5, 7), (−1, 3) (iii) (a, b), (− a, − b)
Answer
(i) Distance between the points is given by
(ii) Distance between (−5, 7) and (−1, 3) is given by
(iii) Distance between (a, b) and (− a, − b) is given by
2. Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2.
Distance between points (0, 0) and (36, 15)
Yes, Assume town A at origin point (0, 0).
Therefore, town B will be at point (36, 15) with respect to town A.
And hence, as calculated above, the distance between town A and B will be 39 km.
Let the points (1, 5), (2, 3), and (- 2,-11) be representing the vertices A, B, and C of the given triangle respectively.Let A = (1, 5), B = (2, 3) and C = (- 2,-11)
Since AB + BC ≠ CA
Therefore, the points (1, 5), (2, 3), and ( – 2, – 11) are not collinear.
Let the points (5, – 2), (6, 4), and (7, – 2) are representing the vertices A, B, and C of the given triangle respectively.
Therefore, AB = BC
As two sides are equal in length, therefore, ABC is an isosceles triangle.
Using distance formula, find which of them is correct.
It can be observed that all sides of this quadrilateral ABCD are of the same length and also the diagonals are of the same length.
Therefore, ABCD is a square and hence, Champa was correct
(i) (- 1, – 2), (1, 0), (- 1, 2), (- 3, 0)
(ii) (- 3, 5), (3, 1), (0, 3), (- 1, – 4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
It can be observed that all sides of this quadrilateral are of the same length and also, the diagonals are of the same length. Therefore, the given points are the vertices of a square.
(ii) Let the points ( – 3, 5), (3, 1), (0, 3), and ( – 1, – 4) be representing the vertices A, B, C, and D of the given quadrilateral respectively.
It can be observed that all sides of this quadrilateral are of different lengths. Therefore, it can be said that it is only a general quadrilateral, and not specific such as square, rectangle, etc.
(iii) Let the points (4, 5), (7, 6), (4, 3), and (1, 2) be representing the vertices A, B, C, and D of the given quadrilateral respectively.
It can be observed that opposite sides of this quadrilateral are of the same length. However, the diagonals are of different lengths. Therefore, the given points are the vertices of a parallelogram.
7. Find the point on the x-axis which is equidistant from (2, – 5) and (- 2, 9).
Answer
We have to find a point on x-axis. Therefore, its y-coordinate will be 0.
Let the point on x-axis be (x,0)
(x – 2)2 + 25 = (x – 2)2 + 81
x2 + 4 – 4x + 25 = x2 + 4 + 4x + 81
8x = 25 -81
8x = -56
x = -7
Therefore, the point is ( – 7, 0).
8. Find the values of y for which the distance between the points P (2, – 3) and Q (10, y) is 10 units.
Answer
It is given that the distance between (2, – 3) and (10, y) is 10.
64 + (y + 3)2 = 100
(y +3)2 = 36
y + 3 = ±6
y + 3 = +6 or y + 3 = -6
Therefore, y = 3 or -9
9. If Q (0, 1) is equidistant from P (5, – 3) and R (x, 6), find the values of x. Also find the distance QR and PR.
Answer
PQ = QR
41 = x2 + 25
16 = x2
x = ±4
Therefore, point R is (4, 6) or ( – 4, 6).
When point R is (4, 6),
10. Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (- 3, 4).
Answer
Point (x, y) is equidistant from (3, 6) and ( – 3, 4).
x2 + 9 – 6x + y2 + 36 – 12y = x2 + 9 + 6x + y2 + 16 – 8y
36 – 16 = 6x + 6x + 12y – 8y
20 = 12x + 4y
3x + y = 5
3x + y – 5 = 0
1. Find the coordinates of the point which divides the join of (- 1, 7) and (4, – 3) in the ratio 2:3.
Answer
Let P(x, y) be the required point. Using the section formula, we get
Therefore, the point is (1, 3).
Answer
3. To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in the following figure. Niharika runs 1/4th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flagexactly halfway between the line segment joining the two flags, where should she post her flag?
Similarly, Preet posted red flag at 1/5 of the distance AD i.e., (1×100/5) m = 20m from the starting point of 8th line. Therefore, the coordinates of this point R are (8, 20).
Distance between these flags by using distance formula = GR
Therefore, Rashmi should post her blue flag at 22.5m on 5th line.
Answer
Let the ratio in which the line segment joining ( -3, 10) and (6, -8) is divided by point ( -1, 6) be k:1.
Therefore, -1 = 6k-3/k+1
–k – 1 = 6k -3
7k = 2
k = 2/7
Therefore, the required ratio is 2:7.
5. Find the ratio in which the line segment joining A (1, – 5) and B (- 4, 5) is divided by the x-axis. Also find the coordinates of the point of division.
Answer
Let the ratio in which the line segment joining A (1, – 5) and B ( – 4, 5) is divided by x-axis be k:1.
Therefore, the coordinates of the point of division is (-4k+1/k+1, 5k-5/k+1).
6. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
Answer
Mid-point of AB is (2, – 3), which is the center of the circle.
⇒ x + 1 = 4 and y + 4 = -6
Therefore, the coordinates of A are (3,-10).
8. If A and B are (–2, –2) and (2, –4), respectively, find the coordinates of P such that AP = 3/7 AB and P lies on the line segment AB.
Answer
10. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2,-1) taken in order. [Hint: Area of a rhombus = 1/2(product of its diagonals)]
Answer
Exercises 7.3
1. Find the area of the triangle whose vertices are:
(i) (2, 3), (-1, 0), (2, -4)
Area of the given triangle = 1/2 [2 { 0- (-4)} + (-1) {(-4) – (3)} + 2 (3 – 0)]
= 1/2 {8 + 7 + 6}
= 21/2 square units.
(ii) Area of the given triangle = 1/2 [-5 { (-5)- (4)} + 3(2-(-1)) + 5{-1 – (-5)}]
= 1/2{35 + 9 + 20}
= 32 square units
2. In each of the following find the value of ‘k‘, for which the points are collinear.
(i) (7, -2), (5, 1), (3, –k)
Answer
(i) For collinear points, area of triangle formed by them is zero.
Therefore, for points (7, -2) (5, 1), and (3, k), area = 0
1/2 [7 { 1- k} + 5(k-(-2)) + 3{(-2) + 1}] = 0
7 – 7k + 5k +10 -9 = 0
-2k + 8 = 0
k = 4
(ii) For collinear points, area of triangle formed by them is zero.
Therefore, for points (8, 1), (k, – 4), and (2, – 5), area = 0
1/2 [8 { -4- (-5)} + k{(-5)-(1)} + 2{1 -(-4)}] = 0
8 – 6k + 10 = 0
6k = 18
k = 3
3. Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.
Answer
Let the vertices of the triangle be A (0, -1), B (2, 1), C (0, 3).
Let D, E, F be the mid-points of the sides of this triangle. Coordinates of D, E, and F are given by
Area of ΔDEF = 1/2 {1(2-1) + 1(1-0) + 0(0-2)}
= 1/2 (1+1) = 1 square units
Area of ΔABC = 1/2 [0(1-3) + 2{3-(-1)} + 0(-1-1)]
= 1/2 {8} = 4 square units
Therefore, the required ratio is 1:4.
4. Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3).
Answer
Let the vertices of the quadrilateral be A ( – 4, – 2), B ( – 3, – 5), C (3, – 2), and D (2, 3). Join AC to form two triangles ΔABC and ΔACD.
Area of ΔABC = 1/2 [(-4) {(-5) – (-2)} + (-3) {(-2) – (-2)} + 3 {(-2) – (-5)}]
= 1/2 (12+0+9)
= 21/2 square units
Area of ΔACD = 1/2 [(-4) {(-2) – (3)} + 3{(3) – (-2)} + 2 {(-2) – (-2)}]
= 1/2 (20+15+0)
= 35/2 square units
Area of ☐ABCD = Area of ΔABC + Area of ΔACD
= (21/2 + 35/2) square units = 28 square units
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 ΔABC whose vertices are A (4, – 6), B (3, – 2) and C (5, 2).
Answer
Let the vertices of the triangle be A (4, -6), B (3, -2), and C (5, 2).
Let D be the mid-point of side BC of ΔABC. Therefore, AD is the median in ΔABC.
= 1/2 (-8+18-16)
= -3 square units
However, area cannot be negative. Therefore, area of ΔABD is 3 square units.
Area of ΔABD = 1/2 [(4) {0 – (2)} + 4{(2) – (-6)} + (5) {(-6) – (0)}]
= 1/2 (-8+32-30)
= -3 square units
However, area cannot be negative. Therefore, area of ΔABD is 3 square units.
The area of both sides is same. Thus, median AD has divided ΔABC in two triangles of equal areas.
Important Links
Coordinate Geometry – Quick Revision Notes
Coordinate Geometry – Most Important Questions
Coordinate Geometry – Important MCQs
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