Ex. 1Multiple choice
If the distance between the points (2, –2) and (–1, x) is 5, one of the values of x is
- (A)–2
- (B)2
- (C)–1
- (D)1
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Solution : Answer (B)
Chapter 7 – Coordinate Geometry
Class 10 Mathematics · 46 questions · 45 with answers
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Ex. 2Multiple choice
The mid-point of the line segment joining the points A (–2, 8) and B (– 6, – 4) is
- (A)(– 4, – 6)
- (B)(2, 6)
- (C)(– 4, 2)
- (D)(4, 2)
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Solution : Answer (C)
Ex. 3Multiple choice
The points A (9, 0), B (9, 6), C (–9, 6) and D (–9, 0) are the vertices of a
- (A)square
- (B)rectangle
- (C)rhombus
- (D)trapezium
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Solution : Answer (B)
Ex. 1Short answerExercise 7.1
The points A (–1, 0), B (3, 1), C (2, 2) and D (–2, 1) are the vertices of a parallelogram.
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Solution : True. The coordinates of the mid-points of both the diagonals AC and BD are ,1 , i.e., the diagonals bisect each other.
Ex. 2Short answerExercise 7.1
The points (4, 5), (7, 6) and (6, 3) are collinear.
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Solution : False. Since the area of the triangle formed by the points is 4 sq. units, the points are not collinear.
Ex. 3Short answerExercise 7.1
Point P (0, –7) is the point of intersection of y-axis and perpendicular bisector of line segment joining the points A (–1, 0) and B (7, –6).
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Solution : True. P (0, –7) lies on the y -axis. It is at a distance of 50 units from both the points (–1, 0) and (7, –6).
Ex. 1Short answerExercise 7.2
If the mid-point of the line segment joining the points A (3, 4) and B (k, 6) is P (x, y) and x + y – 10 = 0, find the value of k. 3+ k 4 + 6
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Solution : Mid-point of the line segment joining A (3, 4) and B (k, 6) = , 2 2 3+ k = ,5 3+ k Then, ,5 = (x, y) 3+ k Therefore, = x and 5 = y. Since x + y – 10 = 0, we have 3+ k + 5 – 10 = 0 i.e., 3 + k = 10 Therefore, k = 7.
Ex. 2Short answerExercise 7.2
Find the area of the triangle ABC with A (1, –4) and the mid-points of sides through A being (2, – 1) and (0, – 1).
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Solution: Let the coordinates of B and C be (a, b) and (x, y), respectively. 1+ a –4 + b Then , , = (2, –1) 2 2 Therefore, 1 + a = 4, –4 + b = –2 a=3 b=2 1+ x –4 + y Also, , = (0, –1) 2 2 Therefore, 1 + x = 0, –4 + y = –2 i.e., x = –1 i.e., y = 2 The coordinates of the vertices of ∆ ABC are A (1, –4), B (3, 2) and C (–1, 2). Area of ∆ ABC = [1(2 – 2)+ 3(2 + 4) –1(– 4 – 2)] = [18 + 6 ] = 12 sq. units.
Ex. 3Short answerExercise 7.2
Name the type of triangle PQR formed by the points P ( 2, 2 ) , ( ) Q – 2, – 2 and R – 6, 6 . ( )
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Solution : Using distance formula 2 2 2 2 PQ = ( 2 + 2 ) + ( 2 + 2 ) = ( 2 2 ) + ( 2 2 ) = 16 = 4 2 2 PR = ( 2 + 6 ) + ( 2 – 6 ) = 2 + 6 + 2 12 + 2 + 6 – 2 12 = 16 = 4 2 2 RQ = ( – 2 + 6 ) + ( – 2 – 6 ) = 2 + 6 − 2 12 + 2 + 6 + 2 12 = 16 = 4 COORDINATE GEOMETRY 83 Since PQ = PR = RQ = 4, points P, Q, R form an equilateral triangle.
Ex. 4Short answerExercise 7.2
ABCD is a parallelogram with vertices A (x1, y1), B (x2, y2) and C (x3, y3). Find the coordinates of the fourth vertex D in terms of x1, x2, x3, y1, y2 and y3.
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Solution: Let the coordinates of D be (x, y). We know that diagonals of a parallelogram bisect each other. x1 + x3 y1 + y3 x + x y2 + y Therefore, mid-point of AC = mid-point of BD , = 2 , 2 2 2 2 i.e., x1 + x3 = x2 + x and y1 + y3 = y2 + y i.e., x1 + x3 – x2 = x and y1 + y3 – y2 = y Thus, the coordinates of D are (x1 + x3 – x2 , y1 + y3 – y2)
Ex. 1Short answerExercise 7.3
The mid-points D, E, F of the sides of a triangle ABC are (3, 4), (8, 9) and (6, 7). Find the coordinates of the vertices of the triangle.
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Solution : Since D and F are the mid-points of AB and AC, respectively, by mid-point theorem, we can prove that DFEB is a parallelogram. Let the coordinates of B be (x, y). Refer to Sample Question 4 of Section (D) to get x=3+8–6=5 y=4+9–7=6 Therefore, B (5, 6) is one of the vertices of the triangle. Similarly DFCE and DAFE are also parallelograms, and the coordinates of A are (3 + 6 – 8, 4 + 7 – 9) = (1, 2). Coordinates of C are (8 + 6 – 3, 9 + 7 – 4) = (11, 12). Thus, the coordinates of the vertices of the triangle are A (1, 2), B (5,6) and C ( 11, 12).
Questions
The distance of the point P (2, 3) from the x-axis is
- (A)2
- (B)3
- (C)1
- (D)5
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(B) 3
The distance between the points A (0, 6) and B (0, –2) is
- (A)6
- (B)8
- (C)4
- (D)2
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(B) 8
The distance of the point P (–6, 8) from the origin is
- (A)8
- (B)2 7
- (C)10
- (D)6
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(C) 10
The distance between the points (0, 5) and (–5, 0) is
- (A)5
- (B)5 2
- (C)2 5
- (D)10
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(B) 5 2
AOBC is a rectangle whose three vertices are vertices A (0, 3), O (0, 0) and B (5, 0). The length of its diagonal is
- (A)5
- (B)3
- (C)34
- (D)4
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(C) 34
The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is
- (A)5
- (B)12
- (C)11
- (D)7 + 5
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(B) 12
The area of a triangle with vertices A (3, 0), B (7, 0) and C (8, 4) is
- (A)14
- (B)28
- (C)8
- (D)6
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(C) 8
The points (–4, 0), (4, 0), (0, 3) are the vertices of a
- (A)right triangle
- (B)isosceles triangle
- (C)equilateral triangle
- (D)scalene triangle COORDINATE GEOMETRY 79
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(B) isosceles triangle
The point which divides the line segment joining the points (7, –6) and (3, 4) in ratio 1 : 2 internally lies in the
- (A)I quadrant
- (B)II quadrant
- (C)III quadrant
- (D)IV quadrant
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(D) IV quadrant
The point which lies on the perpendicular bisector of the line segment joining the points A (–2, –5) and B (2, 5) is
- (A)(0, 0)
- (B)(0, 2)
- (C)(2, 0)
- (D)(–2, 0)
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(A) (0, 0)
The fourth vertex D of a parallelogram ABCD whose three vertices are A (–2, 3), B (6, 7) and C (8, 3) is
- (A)(0, 1)
- (B)(0, –1)
- (C)(–1, 0)
- (D)(1, 0)
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(B) (0, –1)
If the point P (2, 1) lies on the line segment joining points A (4, 2) and B (8, 4), then
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(D)
1 1
- (A)AP = AB
- (B)AP = PB
- (C)PB = AB
- (D)AP = AB
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(B) AP = PB
3 2 13. If P , 4 is the mid-point of the line segment joining the points Q (– 6, 5) and R (– 2, 3), then the value of a is
This question refers to a figure in the original PDF.
- (A)– 4
- (B)– 12
- (C)12
- (D)– 6 14. The perpendicular bisector of the line segment joining the points A (1, 5) and B (4, 6) cuts the y-axis at (A) (0, 13) (B) (0, –13) (C) (0, 12) (D) (13, 0) 15. The coordinates of the point which is equidistant from the three verti- ces of the ∆ AOB as shown in the Fig. 7.1 is (A) (x, y) (B) (y, x) x y y x (C) , (D) , 2 2 2 2 16. A circle drawn with origin as the centre passes through ( ,0) . The point which does not lie in the interior of the circle is –3 7 –1 5 (A) ,1 (B) 2, (C) 5, (D) −6,
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(C) 12
3 2 2 17. A line intersects the y-axis and x-axis at the points P and Q, respectively. If (2, –5) is the mid-point of PQ, then the coordinates of P and Q are, respectively
- (A)(0, – 5) and (2, 0)
- (B)(0, 10) and (– 4, 0)
- (C)(0, 4) and (– 10, 0)
- (D)(0, – 10) and (4, 0) 18. The area of a triangle with vertices (a, b + c), (b, c + a) and (c, a + b) is (A) (a + b + c)2 (B) 0 (C) a + b + c (D) abc 19. If the distance between the points (4, p) and (1, 0) is 5, then the value of p is (A) 4 only (B) ± 4 (C) – 4 only (D) 0 20. If the points A (1, 2), O (0, 0) and C (a, b) are collinear, then (A) a = b (B) a = 2b (C) 2a = b (D) a = –b
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(B) (0, 10) and (– 4, 0)
∆ ABC with vertices A (–2, 0), B (2, 0) and C (0, 2) is similar to ∆ DEF with vertices D (–4, 0) E (4, 0) and F (0, 4). COORDINATE GEOMETRY 81
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True. Because all three sides of both triangles are proportional.
Point P (– 4, 2) lies on the line segment joining the points A (– 4, 6) and B (– 4, – 6).
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True. The three points lie on the line x = –
The points (0, 5), (0, –9) and (3, 6) are collinear.
Point P (0, 2) is the point of intersection of y–axis and perpendicular bisector of line segment joining the points A (–1, 1) and B (3, 3).
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3. False, since two points lie on the y – axis and one point lies in quadrant I. 4. False. PA= 2 and PB= 10 , i.e., PA PB.
Points A (3, 1), B (12, –2) and C (0, 2) cannot be the vertices of a triangle.
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True, since ar (ΔABC) = 0.
Points A (4, 3), B (6, 4), C (5, –6) and D (–3, 5) are the vertices of a parallelo- gram.
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False, since the diagonals donot bisect each other.
A circle has its centre at the origin and a point P (5, 0) lies on it. The point Q (6, 8) lies outside the circle.
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True, radius of the circle = 5 and OP > 5
The point A (2, 7) lies on the perpendicular bisector of line segment joining the points P (6, 5) and Q (0, – 4).
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False, since AP AQ
Point P (5, –3) is one of the two points of trisection of the line segment joining the points A (7, – 2) and B (1, – 5).
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True, since P divides AB in the ratio 1 : 2
Points A (–6, 10), B (–4, 6) and C (3, –8) are collinear such that AB = AC .
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True, since B divides AC in the ratio 2 : 7
The point P (–2, 4) lies on a circle of radius 6 and centre C (3, 5).
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False, since PC = 26 6 , P will lie inside the circle.
The points A (–1, –2), B (4, 3), C (2, 5) and D (–3, 0) in that order form a rectangle.
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True, Mid-points of both the diagonals are the same and the diagonals are of equal length.
Name the type of triangle formed by the points A (–5, 6), B (–4, –2) and C (7, 5).
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Scalene triangle
Find the points on the x–axis which are at a distance of 2 5 from the point (7, –4). How many such points are there?
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(9, 0), (5, 0), 2 points
What type of a quadrilateral do the points A (2, –2), B (7, 3), C (11, –1) and D (6, –6) taken in that order, form?
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Rectangle
Find the value of a , if the distance between the points A (–3, –14) and B (a, –5) is 9 units.
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a = –3
Find a point which is equidistant from the points A (–5, 4) and B (–1, 6)? How many such points are there?
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(–3, 5) the middle point of AB. Infinite number of points. In fact all points which are solutions of the equation 2x+y +1 = 0. –1 19
Find the coordinates of the point Q on the x–axis which lies on the perpendicular bisector of the line segment joining the points A (–5, –2) and B(4, –2). Name the type of triangle formed by the points Q, A and B.
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,0 , isosceles triangle
Find the value of m if the points (5, 1), (–2, –3) and (8, 2m ) are collinear.
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2 14 tt ©
If the point A (2, – 4) is equidistant from P (3, 8) and Q (–10, y), find the values of y. Also find distance PQ.
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y = – 3, – 5, PQ = 290, 13 2
Find the area of the triangle whose vertices are (–8, 4), (–6, 6) and (–3, 9).
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0 –34
In what ratio does the x–axis divide the line segment joining the points (– 4, – 6) and (–1, 7)? Find the coordinates of the point of division. 3 5
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6:7, ,0
Find the ratio in which the point P , divides the line segment joining the 4 12
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1:5
3 points A , and B (2, –5).
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Scalene triangle
2 12. If P (9a – 2, –b) divides line segment joining A (3a + 1, –3) and B (8a, 5) in the ratio 3 : 1, find the values of a and b. 13. If (a, b) is the mid-point of the line segment joining the points A (10, –6) and B (k, 4) and a – 2b = 18, find the value of k and the distance AB. 14. The centre of a circle is (2a, a – 7). Find the values of a if the circle passes through the point (11, –9) and has diameter 10 2 units. 15. The line segment joining the points A (3, 2) and B (5,1) is divided at the point P in the ratio 1:2 and it lies on the line 3x – 18y + k = 0. Find the value of k. –1 5 7 7 16. If D , , E (7, 3) and F , are the midpoints of sides of ∆ ABC, find 2 2 2 2 the area of the ∆ ABC. 17. The points A (2, 9), B (a, 5) and C (5, 5) are the vertices of a triangle ABC right angled at B. Find the values of a and hence the area of ∆ABC. 18. Find the coordinates of the point R on the line segment joining the points P (–1, 3) and Q (2, 5) such that PR = PQ . 19. Find the values of k if the points A (k + 1, 2k), B (3k, 2k + 3) and C (5k – 1, 5k) are collinear. 20. Find the ratio in which the line 2x + 3y – 5 = 0 divides the line segment joining the points (8, –9) and (2, 1). Also find the coordinates of the point of division. COORDINATE GEOMETRY 85
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(9, 0), (5, 0), 2 points
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.
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0, 3 –4 3
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.
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sq. units. x2 x3 y2 y3 x1 x2 x 3 y1 y2 y3 , ,
The points A (x1, y1), B (x2, y2) and C (x3 y3) 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?
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(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?
If the points A (1, –2), B (2, 3) C (a, 2) and D (– 4, –3) form a parallelogram, find the value of a and height of the parallelogram taking AB as base.
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a = –3, h
Students of a school are standing in rows and columns in their playground for a drill practice. A, B, C and D are the positions of four students as shown in figure 7.4. Is it possible to place Jaspal in the drill in such a way that he is equidistant from each of the four students A, B, C and D? If so, what should be his position?
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Yes, Jaspal should be placed at the point (7, 5)
Ayush starts walking from his house to office. Instead of going to the office directly, he goes to a bank first, from there to his daughter’s school and then reaches the office. What is the extra distance travelled by Ayush in reaching his office? (Assume that all distances covered are in straight lines). If the house is situated at (2, 4), bank at (5, 8), school at (13, 14) and office at (13, 26) and coordinates are in km.
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House to Bank = 5 km Bank to school = 10 km School to Office = 12 km Total distance travelled = 27 km Distance from house to office = 24.6 km Extra distance = 2.4 km tt ©