During revision, students quickly go through Class 9 Maths Important Questions and Ganita Manjari Class 9 Maths Chapter 1 Orienting Yourself The Use of Coordinates Important Questions with Solutions for clarity.
Class 9 Orienting Yourself The Use of Coordinates Important Questions
Important Questions of Orienting Yourself The Use of Coordinates
Very Short Answer Type Questions
Question 1.
M is a point on the y-axis at a distance of 4 units from the x-axis, and it lies below the x-axis. Find the distance of point M from point Q(5, 1).
Solution:
Since M lies on the y-axis, its x-coordinate = 0
Also, it lies 4 units below the x-axis, so its y-coordinate = -4
Therefore, the coordinates of the point M(0, -4)
The coordinates of the other point are Q(5, 1)
Using the Baudhayana-Pythagoras Theorem, we get
MQ = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
= \(\sqrt{(5-0)^2+(1-(-4))^2}\)
= \(\sqrt{(5)^2+(1+4)^2}\)
= \(\sqrt{(5)^2+(5)^2}\)
= \(\sqrt{25+25}\)
= √50 units
So, the required distance of M from point Q is √50 units.
Question 2.
Find the coordinates of a point on the x-axis which is equidistant from the points whose coordinates are (2, 1) and (3, 5).
Solution:
Any point on the x-axis is of the form P(x, 0).
Let this point be at an equal distance from the points A(2, 1) and B(3, 5).
Thus, PA = PB
⇒ \(\sqrt{(x-2)^2+(0-1)^2}=\sqrt{(x-3)^2+(0-5)^2}\)
⇒ \(\sqrt{\left(x^2-4 x+4\right)^2+(-1)^2}=\sqrt{\left(x^2-6 x+9\right)+(-5)^2}\)
Squaring both sides, we get
⇒ x2 – 4x + 4 + 1 = x2 – 6x + 9 + 25
⇒ 2x = 29
⇒ x = \(\frac {29}{2}\)
So the required point on the x-axis is (\(\frac {29}{2}\), 0)
Question 3.
End points of the diameter of a circle lie at points (-4, 4) and (4, 2). Its centre lies on the y-axis. Find the coordinates of its centre.
Solution:
Let the given end points of the diameter be represented by A(-4, 4) and B(4, 2).
Let its centre, which lies on the y-axis, be represented by O(0, y).
In a circle, the centre is at equal distances from the points lying on the circle.
Hence, OA = OB
⇒ \(\sqrt{(0-(-4))^2+(y-4)^2}=\sqrt{(0-4)^2+(y-2)^2}\)
Squaring both sides, we obtain
(0 + 4)2 + (y – 4)2 = (-4)2 + (y – 2)2
⇒ 16 + y2 – 8y + 16 = 16 + y2 – 4y + 4
⇒ -4y = -12
⇒ y = 3
So, required coordinates of the centre are O(0, 3).
Class 9 Maths Orienting Yourself The Use of Coordinates Important Questions
Short Answer Type Questions
Question 1.
Show that the points (-3, -3),(3, 3) and (-3√3, 3√3) are the vertices of an equilateral triangle.
Solution:
Let given points be represented by A(-3, -3), B(3, 3) and C(-3√3, 3√3).
To show these points as the vertices of an equilateral triangle, we must show that AB = BC = CA.
Now, AB = \(\sqrt{(3-(-3))^2+(3-(-3))^2}\)
= \(\sqrt{(3+3)^2+(3+3)^2}\)
= \(\sqrt{6^2+6^2}\)
= \(\sqrt{2 .(6)^2}\)
= 6√2 units …..(i)
BC = \(\sqrt{(-3 \sqrt{3}-3)^2+(3 \sqrt{3}-3)^2}\)
= \(\sqrt{27+18 \sqrt{3}+9+27-18 \sqrt{3}+9}\)
= √72
= \(\sqrt{2 .(6)^2}\)
= 2√6 units …..(ii)
CA = \(\sqrt{(-3 \sqrt{3}-(-3))^2+(3 \sqrt{3}-(-3))^2}\)
= \(\sqrt{(-3 \sqrt{3}+3)^2+(3 \sqrt{3}+3)^2}\)
= \(\sqrt{27-18 \sqrt{3}+9+27+18 \sqrt{3}+9}\)
= √72
= \(\sqrt{2 .(6)^2}\)
= 6√2 units ……(iii)
From equations (i), (ii), and (iii), we can conclude that AB = BC = CA.
Since the sides of a triangle are all equal,
So, the given points are the vertices of an equilateral triangle.
Orienting Yourself The Use of Coordinates Important Questions Class 9
Long Answer Type Questions
Question 1.
Prove that A(4, 3), B(6, 4), C(5, 6), D(3, 5) are the vertices of a square ABCD.
Solution:
To show that A(4, 3), B(6, 4), C(5, 6), D(3, 5) are the vertices of a square ABCD, we find the following distances:
AB = \(\sqrt{(6-4)^2+(4-3)^2}\)
= \(\sqrt{2^2+1^2}\)
= \(\sqrt{4+1}\)
= √5 units
BC = \(\sqrt{(5-6)^2+(6-4)^2}\)
= \(\sqrt{(-1)^2+2^2}\)
= \(\sqrt{4+1}\)
= √5 units
CD = \(\sqrt{(3-5)^2+(5-6)^2}\)
= \(\sqrt{(-2)^2+(-1)^2}\)
= \(\sqrt{4+1}\)
= √5 units
DA = \(\sqrt{(4-3)^2+(3-5)^2}\)
= \(\sqrt{1^2+(-2)^2}\)
= \(\sqrt{1+4}\)
= √5 units,
AC = \(\sqrt{(5-4)^2+(6-3)^2}\)
= \(\sqrt{1^2+3^2}\)
= \(\sqrt{1+9}\)
= √10 units
BD = \(\sqrt{(3-6)^2+(5-4)^2}\)
= \(\sqrt{(-3)^2+1^2}\)
= \(\sqrt{9+1}\)
= √10 units
So, AB = BC = CD + DA and AC = BD
All four sides are equal, and both diagonals are equal.
Therefore, A, B, C, D are the vertices of a square.
Proved.
Class 9 Maths Orienting Yourself The Use of Coordinates Important Questions
Case-Study Based Questions
Question 1.
Trees act as natural filters. By planting trees in and around school premises, we create cleaner and healthier air for the students and residents, reducing respiratory problems. On World Earth Day, which falls every year on April 22, Kendriya Vidyalaya in Godda had proposed and organised a community drive on tree plantation under the title “Plant Trees for Generations to Come”. Students of that school planted saplings in the field such that it formed a quadrilateral as shown in the figure ABCD.
Based on the information given above, answer the following questions:
(i) Write the coordinates of the points A, B, C and D.
(ii) Find the distance between the two saplings at A and D.
(iii) Find the distances between AC and BD. Are they equal?
(iv) Find the distances between PB and PD. Can we conclude that P is the midpoint of the diagonal BD?
Solution:
(i) Coordinates of the required points are A(-5, 9), B(2, 3), C(1, -5), D(-6, 1).
(ii) Distance between and A(-5, 9) and D(-6, 1) is given by the Baudhayana-Pythagoras Formula as
AD = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
= \(\sqrt{6-(-5)^2+(1-9)^2}\)
= \(\sqrt{(-6+5)^2+(-8)^2}\)
= \(\sqrt{(-1)^2+(-8)^2}\)
= \(\sqrt{(1+64)}\)
= √65
(iii) Using the above formula, we get the distance between A(-5, 9) and C(1, -5) as
AC = \(\sqrt{\left[(1-(-5)]^2+(-5-9)^2\right.}\)
= \(\sqrt{(1+5)^2+(-14)^2}\)
= \(\sqrt{6^2+(-14)^2}\)
= \(\sqrt{(36+196)}\)
= √232 units
And BD = \(\sqrt{(-6-2)^2+(1-3)^2}\)
= \(\sqrt{(-8)^2+(-2)^2}\)
= \(\sqrt{(64+4)}\)
= √68 units.
Clearly, AC ≠ BD.
(iv) Using the distance formula, already used above, we can find the distance between P(-2, 2) and B(2, 3) as
PB = \(\sqrt{\left[(2-(-2)]^2+(3-2)^2\right.}\)
= \(\sqrt{(2+2)^2+(1)^2}\)
= \(\sqrt{(-4)^2+(-1)^2}\)
= \(\sqrt{16+1}\)
= √17 units
And, that between P(-2, 2) and D(-6, 1) as
PD = \(\sqrt{\left[(-6-(-2)]^2+(1-2)^2\right.}\)
= \(\sqrt{(-6+2)^2+(-1)^2}\)
= \(\sqrt{(-4)^2+(-1)^2}\)
= \(\sqrt{16+1}\)
= √17 units
Clearly, PB = PD.
Yes, we can conclude that P is the midpoint of B and D as the distances of P from B and D are equal and they are on the same line segment BD.
Question 2.
In a society, there is a circular park having two gates. The gates are placed at points A(10, 20) and B(50, 50), as shown in the figure below.
Two fountains are installed at points P and Q on AB. Coordinates of points are P(\(\frac {70}{3}\), 30), C(30, 35), and Q(\(\frac {110}{3}\), 40).
Based on the above information, answer the following questions:
(i) Find the distances between A and C, and B and C. Are they equal?
(ii) Find the distances between P and C, and Q and C. Are they equal?
(iii) Find the distance between P and Q, and A and P. Are they equal?
(iv) Find the distance between B and Q. Also, find the ratio of AP : PQ : QB.
Answer:
(i) AC = \(\sqrt{(30-10)^2+(35-20)^2}\)
= \(\sqrt{(20)^2+(15)^2}\)
= \(\sqrt{400+225}\)
= √625
= 25 units
BC = \(\sqrt{(50-30)^2+(50-35)^2}\)
= \(\sqrt{(20)^2+(15)^2}\)
= \(\sqrt{400+225}\)
= √625
= 25 units
Yes, AC = BC.
Yes, PC = QC.
Yes, PQ = AP.
Since, AP = \(\frac {50}{3}\) units, PQ = \(\frac {50}{3}\) units and QB = \(\frac {50}{3}\) units.
Hence, AP : PQ : QB = \(\frac{50}{3}: \frac{50}{3}: \frac{50}{3}\)
⇒ AP : PQ : QB = 1 : 1 : 1.
Self Assessment
Multiple Choice Questions (MCQ)
Question 1.
In the coordinates of the point A(2, -4), the abscissa is:
(a) 0
(b) -1
(c) 2
(d) 3
Answer:
(c) 2
Question 2.
In the coordinates of the point P(-2, 1), the ordinate is:
(a) 3
(b) 1
(c) -2
(d) 0
Answer:
(b) 1
Question 3.
A point has its abscissa equal to 2 and ordinate equal to -3; the coordinates of the point are:
(a) (2, -3)
(b) (-2, 3)
(c) (-2, -3)
(d) (-3, 2)
Answer:
(a) (2, -3)
Question 4.
A point has x-coordinate equal to -1 and y-coordinate equal to zero. In which quadrant or on which axis does this point lie?
(a) x-axis
(b) y-axis
(c) second quadrant
(d) fourth quadrant
Answer:
(a) x-axis
Question 5.
A point has an x-coordinate equal to zero and a y-coordinate equal to -9. In which quadrant or on which axis does this point lie?
(a) x-axis
(b) y-axis
(c) second quadrant
(d) fourth quadrant
Answer:
(b) y-axis
Question 6.
In which quadrant does the point (1, 2) lie?
(a) Ist Quadrant
(b) IInd Quadrant
(c) IIIrd Quadrant
(d) IVth Quadrant
Answer:
(a) Ist Quadrant
Question 7.
In which quadrant does the point (-2, -7) lie?
(a) Ist Quadrant
(b) IInd Quadrant
(c) IIIrd Quadrant
(d) IVth Quadrant
Answer:
(c) IIIrd Quadrant
Question 8.
In which quadrant does the point (5, -1) lie?
(a) Ist Quadrant
(b) IInd Quadrant
(c) IIIrd Quadrant
(d) IVth Quadrant
Answer:
(d) IVth Quadrant
Question 9.
In which quadrant does the point (-5, 2) lie?
(a) Ist Quadrant
(b) IInd Quadrant
(c) IIIrd Quadrant
(d) IVth Quadrant
Answer:
(b) IInd Quadrant
Question 10.
A point lies below the x-axis and to the left of the y-axis. In which quadrant does this point lie?
(a) Ist Quadrant
(b) IInd Quadrant
(c) IIIrd Quadrant
(d) IVth Quadrant
Answer:
(c) IIIrd Quadrant
Assertion-Reason Based Questions
Direction: A statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option from the following options.
(a) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true, and Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) is false, but Reason (R) is true.
Question 1.
Assertion (A): For the point (1, -9), the value of (abscissa-ordinate) = 10.
Reason (R): In the coordinates of any point (x, y), x is called abscissa, and y is called ordinate.
Answer:
(a) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).
Question 2.
Assertion (A): The points (0, 0),(0, -2) and (0, 7) are collinear points.
Reason (R): Any point of the form (±a, 0) lies on the x-axis.
Answer:
(b) Both Assertion (A) and Reason (R) are true, and Reason (R) is not the correct explanation of Assertion (A).
Question 3.
Assertion (A): Points (-4, 2), (-1, 6), (-2, 4), and (-5, 1) lie in different quadrants.
Reason (R): Points of the form (-u, v) lie in the IInd Quadrant.
Answer:
(d) Assertion (A) is false, but Reason (R) is true.
Question 4.
Assertion (A): If an object is placed at a point (-1, 1) and a mirror is placed along the y-axis, then its image will be at the point (1, 1).
Reason (R): The image of a point (a, b) in the mirror along the x-axis is given by the point (a, -b).
Answer:
(b) Both Assertion (A) and Reason (R) are true, and Reason (R) is not the correct explanation of Assertion (A).
Question 5.
Assertion (A): If A(3, -3) and B(4, -3), then AB = 1 unit.
Reason (R): The distance between the points of the form (u, v) and (w, v) is given by |w – u|.
Answer:
(a) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).
Question 6.
Assertion (A): If U(3, -7) and V(-2, -2), then UV = 5√2 units.
Reason (R): The distance between the points of the forms (u, v) and (w, x) is given by \(\sqrt{(w-u)^2+(x-v)^2}\) units.
Answer:
(a) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).
Case Study Based Questions
Question 1.
Aarya was doing her Maths homework. Her sister Aadya was sitting beside her. When Aarya was busy doing her homework, Aadya took a graph sheet out of Aarya’s bag and put four points on the graph sheet as shown in the following figure. Aarya had already plotted axes on it and had already marked points on the axes.
Based on the information given above, answer the following questions:
(i) Write the coordinates of the points R, E, S, and T.
(ii) Find the distances between R and E, S, and T.
(iii) Find the distances between R and T, S and E.
(iv) What geometrical figure do you get by joining REST?
Answer:
(i) R(-4, 2), E(1, 2), S(4, -2), T(-6, -2).
(ii) RE = 5 units, ST = 10 units.
(iii) RT = 2√5 units, SE = 5 units.
(iv) Trapezium
Very Short Answer Type Questions
Question 1.
M is a point on the y-axis at a distance of 2 units from the x-axis, and it lies above the x-axis. Find the distance of point M from point Q(-2, 1).
Answer:
√5 units
Question 2.
Find the midpoint of (-2, 4) and (5, -4). In which quadrant or on which axis does this point lie?
Answer:
(\(\frac {3}{2}\), 0), x-axis
Question 3.
Find the distance between the points A(3, -3) and (a + 3, a – 3).
Answer:
a√2 units
Question 4.
Find the reflection of the point P(7, -4) in the mirror placed along the y-axis.
Answer:
(-7, -4)
Question 5.
Find the distance of the point Z(√5, -√3) from the origin.
Answer:
√8 units
Short Answer Type Questions
Question 1.
Show that the points A(-3, -3), B(3, 3) and C(5, 5) are collinear.
Answer:
Show AB + BC = AC
Question 2.
Show that P(2, -3) is the mid-point of the line segment joining the points U(6, -9) and V(-2, 3).
Answer:
Show PU = PV
Question 3.
Find a point on the y-axis which is equidistant from the points (3, 1) and (4, 8).
Answer:
(0, 5)
Long Answer Type Questions
Question 1.
Prove that A(0, 1), B(2, 4), C(6, 4), D(4, 1) are the vertices of a parallelogram ABCD but not a rectangle ABCD.
Answer:
Prove that both pairs of opposite sides are equal, but its diagonals are not equal.
Question 2.
Show that the points (1, 0), (-1, 0) and (0, 1) lie on a circle whose centre lies at the origin.
Answer:
Show that the distances of the points (1, 0), (-1, 0) and (0, 1) from the origin are equal.
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