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MCQ on Orienting Yourself The Use of Coordinates Class 9
Class 9 Maths Orienting Yourself The Use of Coordinates MCQ
Choose the correct option from the given options:
Question 1.
In the coordinates of the point A(-1, 2), the abscissa is :
(a) 0
(b) -1
(c) 2
(d) 3
Solution:
(b) -1
Explanation:
The first component i.e., x-coordinate, in the coordinates of a point is called abscissa of the point.
Question 2.
In the coordinates of the point P(3, -2), the ordinate is :
(a) 3
(b) 1
(c) -2
(d) 0
Solution:
(c) -2
Explanation:
The second component i.e., y—coordinate, in the coordinates of a point is called ordinate of the point.
Question 3.
A point has its abscissa equals to 4 and ordinate equals to -1, the coordinates of the point is :
(a) (4, -1)
(b) (-1, 4)
(c) (4, 0)
(d) (0, -1)
Solution:
(a) (4, -1)
Explanation:
Abscissa is the x-coordinate and ordinate is the ^-coordinate in the coordinates of the point.
Question 4.
A point has x-coordinate equals to -4 and y-coordinate equals 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
Solution:
(a) x-axis
Explanation:
Any point of the form (± a, 0) lies on the x-axis. Here, the point is of the form (—a, 0), so the point (-4, 0) lies on the x-axis.
Question 5.
A point has x-coordinate equals to zero and y- coordinate equals to -11. In which quadrant or on which axis does this point lie?
(a) x-axis
(b) y-axis
(c) second quadrant
(d) fourth quadrant
Solution:
(b) y-axis
Explanation:
Any point of the form (0,± a) lies on the y-axis. Here, the point is of the form (0, – a), so the point (0, -11) lies on the y-axis.
Orienting Yourself The Use of Coordinates MCQ Class 9
Question 6.
In which quadrant does the point (-1, -2) lie?
(a) Ist Quadrant
(b) IInd Quadrant
(c) IIIrd Quadrant
(d) IVth Quadrant
Solution:
(c) IIIrd Quadrant
Explanation:
Any point in the third quadrant is of the form (-, -). Given point is in the form (-, -), so (—1, -2) lies in the IIIrd quadrant.
Question 7.
In which quadrant does the point (-3, 7) lie?
(a) Ist Quadrant
(b) IInd Quadrant
(c) IIIrd Quadrant
(d) IVth Quadrant
Solution:
(b) IInd Quadrant
Explanation:
Any point in the second quadrant is of the form (—, +). Given point is in the form (-, +), so (-3, 7) lies in the IInd quadrant.
Question 8.
In which quadrant does the point (15,-21) lie?
(a) Ist Quadrant
(b) IInd Quadrant
(c) IIIrd Quadrant
(d) IVth Quadrant
Solution:
(d) IVth Quadrant
Explanation:
Any point in the fourth quadrant is of the form (+, —). Given point is in the form (+, —), so (15, —21) lies in the IVth quadrant.
Question 9.
In which quadrant does the point (15,12) lie?
(a) Ist Quadrant
(b) IInd Quadrant
(c) IIIrd Quadrant
(d) IVth Quadrant
Solution:
(a) Ist Quadrant
Explanation:
Any point in the first quadrant is of the form (+, +). Given point is in the form (+, +), so (15, 12) lies in the Ist quadrant.
Question 10.
A point lies above x-axis and to the right of y-axis. In which quadrant does this point lie?
(a) Ist Quadrant
(b) IInd Quadrant
(c) IIIrd Quadrant
(d) IVth Quadrant
Solution:
(a) Ist Quadrant
Explanation:
Any point which lies to the right of y-axis will have positive x—coordinate. Also, any point which lies above x-axis will have positive y— coordinate. So, the point in the question has coordinates of the form (+, +). Therefore, it will lie in the Ist quadrant.
Question 11.
A point lies below x-axis and to the left of y-axis. In which quadrant does this point lie?
(a) Ist Quadrant
(b) IInd Quadrant
(c) IIIrd Quadrant
(d) IVth Quadrant
Solution:
(a) Ist Quadrant
Explanation:
Any point which lies to the left of y-axis will have negative x-coordinate. Also, any point which lies below x-axis will have negative y-coordinate. So, the point in the question has coordinates of the form (-, -). Therefore, it will lie in the IIIrd quadrant.
Question 12.
A point lies in the second quadrant. In which quadrant does its reflection lie, if the mirror is placed along the x-axis?
(a) Ist Quadrant
(b) IInd Quadrant
(c) IIIrd Quadrant
(d) IVth Quadrant
Solution:
(c) IIIrd Quadrant
Explanation:
Any point which lies in the second quadrant will have coordinates of the form (-, +). If a mirror is placed along the x-axis, then its reflection will have same sign for the x-coordinate but opposite sign for the y— coordinate. So, coordinates of its reflection will be of the form (-, -) suggesting us that it will lie in the IIIrd quadrant.
Question 13.
A point lies in the third quadrant. In which quadrant does its reflection lie, if the mirror is placed along the y-axis?
(a) Ist Quadrant
(b) IInd Quadrant
(c) IIIrd Quadrant
(d) IVth Quadrant
Solution:
(d) IVth Quadrant
Explanation:
Any point which lies in the third quadrant will have coordinates of the form (—, -). If a mirror is placed along the y-axis, then its reflection will have same sign for the y-coordinate but opposite sign for the x— coordinate. So, coordinates of its reflection will be of the form (+, —) suggesting us that it will lie in the IVth quadrant.
Question 14.
Point P has x-coordinate equal to -7. The coordinates of point Q which is on the line through P parallel to the y-axis and at a distance of 3 units below the x-axis are :
(a) (-7, 3)
(b) (-3, -7)
(c) (-7, -3)
(d) (-3,-7)
Solution:
(c) (-7, -3)
Explanation:
Any point on the line parallel to the y-axis will have same x-coordinate. So, point Q will have x-coordinate equal to -7. Also, any point which is below the x-axis will have negative y-coordinate equal to the units it is below the x-axis. So, Q will have its y-coordinate equal to -3, as it is 3 units below the x-axis. Finally, we can write the coordinates of the point as Q(-7, -3).
Question 15.
Point R has y-coordinate equal to 11. The coordinates of point S which is on the line through R parallel to the x-axis and at a distance of 5 units to the right of the y-axis are :
(a) (11, -5)
(b) (-11,-5)
(c) (5, -11)
(d) (5, 11)
Solution:
(d) (5, 11)
Explanation:
Any point on the line parallel to the x-axis will have same y-coordinate. So, point S will have y-coordinate equal to 11. Also, any point which is to the right of the y-axis will have positive x-coordinate equal to the units it is right of the y-axis. So, S will have its x- coordinate equal to 5, as it is 5 units to the right of the y-axis. Finally, we can write the coordinates of the point as S(5, 11).
Question 16.
The shortest distance (in units) of the point (2, 3) from y-axis is :
(a) 2
(b) 3
(c) 5
(d) 1
Solution:
(a) 2
Explanation:
The shortest distance of any point from the y-axis is given by the absolute value of its x— coordinate, i.e., the number without negative sign if it has a negative sign. Here, the given point is (2, 3), whose x-coordinate is 2. So, its distance from the y-axis is 2 units.
Question 17.
The shortest distance (in units) of the point (-4, -7) from y-axis is :
(a) -4
(b) 4
(c) -7
(d) \(\sqrt{65}\)
Solution:
(b) 4
Explanation:
The shortest distance, of any point from the y-axis is given by the absolute value of its x— coordinate, i.e., the number without negative sign if it has a negative sign. Here, the given point is (-4, —7), whose x—coordinate is -4. Absolute value of -4 is |— 4| = 4. So, its distance from the y-axis is 4 units.
Question 18.
The shortest distance (in units) of the point (2, -5) from x-axis is :
(a) \(\sqrt{27}\)
(b) 2
(c) -5
(d) 5
Solution:
(d) 5
Explanation:
The shortest distance of any point from the x-axis is given by the absolute value of its y—coordinate, i.e., the number without negative sign if it has a negative sign. Here, the given point is (2,-5), whose y-coordinate is -5. Absolute value of -5 is |—5| = 5. So, its distance from the y-axis is 5 units.
Question 19.
Points A(-12, 0), B(-4, 0) and C(6, 0) are :
(a) vertices of a right-angled triangle
(b) vertices of a scalene triangle
(c) vertices of an equilateral triangle
(d) collinear
Solution:
(d) collinear
Explanation:
Given points A(-12, 0), B(-4, 0) and C(6, 0) lie on the x-axis, so these points are collinear points. Three or more points are said to he collinear if they lie on the same line.
Question 20.
Points A(0, 2),B(-4, 0) and C(0, 0) are :
(a) vertices of a right-angled triangle
(b) vertices of a scalene triangle
(c) vertices of an equilateral triangle
(d) collinear
Solution:
(c) vertices of an equilateral triangle
Explanation:
Given points A(0, 2), B(-4, 0) and C(0, 0) lie on the y-axis, x-axis and at the origin, respectively. So, by joining these points we 4 get a right-angled triangle. Therefore, these points are vertices of a right-angled triangle.
Question 21.
The distance between the points (2, 3) and (-1, -1) is :
(a) \(\sqrt{13}\) units
(b) \(\sqrt{15}\) units
(c) 5 units
(d) -5 units
Solution:
(c) 5 units
Explanation:
Distance between the points A(x1, y2) and B(x2, y1) is given by the Baudhayana— Pythagoras Theorem as
AB = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
Hence, the distance between the given points is given by \(\sqrt{100}\)
= \(\sqrt{100}\)
= 5 units.
Question 22.
The distance between the points (12, 5) and (0, 0) is :
(a) 17 units
(b) 15 units
(c) 13 units
(d) -13 units
Solution:
(c) 13 units
Explanation:
Distance between the points A(x1, y2) and B(x2, y1) is given by the Baudhayana- Pythagoras Theorem as
AB = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
Hence, the distance between the given points is given by \(\frac{1}{2}\)
= \(\frac{1}{2}\)
= 13 units.
Question 23.
The distance between the points (12, 0) and (-3, 0) is :
(a) 9 units
(b) 15 units
(c) 12 units
(d) -3 units
Solution:
(b) 15 units
Explanation:
Distance between the points A(x1, y2) and B(x2, y1) is given by AB = | x2 — x: |.
Hence, the distance between the points (12, 0) and (-3, 0) is given by |x2 – x1|
= |12 – (-3)|
= |12 + 3|
= |15|
= 15 units.
Question 24.
The distance between the points (0, -2) and (0, -13) is :
(a) 9 units
(b) 15 units
(c) 11 units
(d) -11 units
Solution:
(c) 11 units
Explanation:
Distance between the points A (0, x1) and B (0, y2) is given by AB = |y2 – y1|.
Hence, the distance between the points (0, -2) and (0, -13) is given by |y2 – y1|
= |-13-(-2)| = |-13 + 2|
= |—11|
= 11 units.
Question 25.
The distance between the points (a – b, a + b) and (a + b, b – a) is :
(a) 2\(\sqrt{a^2+b^2}\) units
(b) \(\sqrt{a^2+b^2}\) units
(c) 2\(\sqrt{a^2+b^2}\) units
(d) 2\(\sqrt{a^2-b^2}\) units
Solution:
(a) 2\(\sqrt{a^2+b^2}\) units
Explanation:
Distance between the points A(a – b, a + b) and B(a + b, b – a) is given by
AB = \(\sqrt{\{(a+b)-(a-b)\}^2+\{(b-a)-(a+b)\}^2}\)
So, AB = \(\sqrt{\{a+b-a+b\}^2+\{b-a-a-b\}^2}\)
⇒ AB = \(\sqrt{\{2 b\}^2+\{-2 a\}^2}\)
⇒ AB = \(\sqrt{4 b^2+4 a^2}\)
⇒ AB = \(\sqrt{4 b^2+a^2}\)
AB = 2\(\sqrt{a^2+b^2}\) units
Question 26.
The mid-point of the line segment joining the points (4, 6) and (2, 2) is :
(a) (3, 3)
(b) (3, 4)
(c) (4, 3)
(d) (1, 2)
Solution:
(b) (3, 4)
Explanation:
The mid-point of a line segment divides it into two equal halves.
Let P(x, y) be the mid—point of the line segment joining the points A(4, 6) and B(2, 2).
So, PA = PB
⇒ \(\sqrt{(4-x)^2+(6-y)^2}=\sqrt{(2-x)^2+(2-y)^2}\)
⇒16 – 8x + x2 + 36 – 12y + y2 = 4 – 4x + x2 + 4 – 4y + y2
⇒ x2 + y2 – 8x – 12y + 52 = x2 + y2 – 4x – 4y + 8
⇒ x2 + y2 – 8x – 12y + 52 – x2 – y2 + 4x + 4y – 8 = 0
⇒ -4x – 8y + 44 = 0
⇒ x + 2y – 11 = 0
Amongst the given points, (3, 4) satisfies the above equation. Hence, required mid-point of the given line segment whose end points are (4, 6) and (2, 2) is the point whose coordinates are (3, 4).
Aliter: To get the mid-point of any line segment whose end points are given, just add their x-coordinates of the points and divide the sum by 2. Do the same for their y- coordinates.
Let us suppose that the given points are (x1, y1) and (x2, y2). The mid-point will be
\(\frac{1}{2}\)
Here, the points are (4, 6) and (2, 2). So, x1 = 4, x2 = 2, y1 = 6, y2 = 2.
The required mid-point can be got as
\(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)=\left(\frac{4+2}{2}, \frac{6+2}{2}\right)\)
= \(\left(\frac{6}{2}, \frac{8}{2}\right)\) = (3, 4)
Question 27.
End points of a diameter of a circle are given by (4, -2) and (6, 8), the centre of the circle will lie at:
(a) (5, 5)
(b) (5, 3)
(c) (-5, 3)
(d) (-5,-3)
Solution:
(b) (5, 3)
Explanation:
For the given points (x1, y1) and (x2, y2) the mid-point will be
\(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)
So, mid-point will be \(\left(\frac{4+6}{2}, \frac{-2+8}{2}\right)\)
= \(\left(\frac{10}{2}, \frac{6}{2}\right)\) = (5, 3)
Question 28.
The points A(9, 0), B(9, -6), C(-9, 0) and D(-9, 6) are the vertices of a
(a) Square
(b) Rectangle
(c) Parallelogram
(d) Trapezium
Solution:
(c) Parallelogram
Explanation:
Here, AB = |y2 – y1| = |—6 – 0| = |—6| = 6.
CD = |y2 – y1| = |6-0| = |6| = 6.
BC = \(\sqrt{(-9-9) 2+(0-(-6))^2}\)
= \(\sqrt{(-18)^2+(6)^2}=\sqrt{324+36}=\sqrt{360}\)
AD = \(\sqrt{(-9-9)^2+(6-0)^2}\)
= \(\sqrt{(-18)^2+(6)^2}=\sqrt{324+36}=\sqrt{360}\)
AC = \(\sqrt{(-9-9)^2+(0-0)^2}=\sqrt{(-18)^2+(0)^2}\)
= \(\sqrt{324+0}=\sqrt{324}\) = 18
BD = \(\sqrt{(-9-9)^2+(6-(-6))^2}\)
= \(\sqrt{(-18)^2+(12)^2}=\sqrt{324+144}=\sqrt{368}\)
Clearly, AB = CD, BC = AD, but AC ≠ BD.
So, given points are the vertices of a parallelogram.
Question 29.
In the coordinates of the point A(2, -4), the abscissa is :
(a) 0
(b) -1
(c) 2
(d) 3
Solution:
(c) 2
Question 30.
In the coordinates of the point P(-2, 1), the ordinate is :
(a) 3
(b) 1
(c) -2
(d) 0
Solution:
(b) 1
Question 31.
A point has its abscissa equals to 2 and ordinate equals to -3, the coordinates of the point is :
(a) (2, -3)
(b) (-2, 3)
(c) (-2, -3)
(d) (-3, 2)
Solution:
(a) (2, -3)
Question 32.
A point has x-coordinate equals to -1 and ^-coordinate equals 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
Solution:
(a) x-axis
Question 33.
A point has x-coordinate equals to zero and y-coordinate equals 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
Solution:
(b) y-axis
Question 34.
In which quadrant does the point (1, 2) lie?
(a) Ist Quadrant
(b) IInd Quadrant
(c) IIIrd Quadrant
(d) IVth Quadrant
Solution:
(a) Ist Quadrant
Question 35.
In which quadrant does the point (-2, -7) lie?
(a) Ist Quadrant
(b) IInd Quadrant
(c) IIIrd Quadrant
(d) IVth Quadrant
Solution:
(c) IIIrd Quadrant
Question 36.
In which quadrant does the point (5, -1) lie?
(a) Ist Quadrant
(b) IInd Quadrant
(c) IIIrd Quadrant
(d) IVth Quadrant
Solution:
(d) IVth Quadrant
Question 37.
In which quadrant does the point (-5, 2) lie?
(a) Ist Quadrant
(b) IInd Quadrant
(c) IIIrd Quadrant
(d) IVth Quadrant
Solution:
(b) IInd Quadrant
Question 38.
A point lies below x-axis and to the left of y-axis. In which quadrant does this point lie?
(a) Ist Quadrant
(b) IInd Quadrant
(c) IIIrd Quadrant
(d) IVth Quadrant
Solution:
(c) IIIrd Quadrant
Orienting Yourself The Use of Coordinates Class 9 Assertion and Reason 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 (3, -5), the value of (absecissa — ordinate) = 8.
Reason (R): In the coordinate of any point (x, y), x is called abscissa and y is called ordinate.
Solution:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Explanation:
Here, for the point abscissa = 3 and ordinate = -5
So, abscissa — ordinate = 3 – (-5) = 3 + 5 = 8.
Thus, Assertion (A) is true.
Also, in any coordinates (x, y), x is its abscissa and y is its ordinate.
Therefore, Reason (R) is true and Reason (R) is a good explanation of Assertion (A) as well.
Question 2.
Assertion (A): For the points (-4, 0), (3, 0) and (5, 0) are collinear points.
Reason (R): Any point of the form (± a, 0) lies on the x-axis.
Solution:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Explanation:
Here, the points (-4, 0), (3, 0) and (5, 0) are of the form (± a, 0).
So, they lie on x-axis.
Thus, Assertion (A) is true.
Also, if any point has coordinates of the form (± a, 0) lies on the x-axis.
Therefore, Reason (R) is true.
Reason (R) is a good explanation of Assertion (A) as points lying on a line are said to be collinear points.
Question 3.
Assertion (A): Points (-4, 2), (1, 6), (-2, -4) and (5, -1) lie in different quadrants.
Reason (R): Points with different values of x and y-coordinates, lie in different quadrants.
Solution:
(c) Assertion (A) is true but Reason (R) is false.
Explanation:
Here, the points (-4, 2), (1, 6), (-2, -4) and (5, -1) lie in different quadrants as (-4, 2) lies in IInd Quadrant, (1, 6) lies in 1st Quadrant, (-2, -4) lies in IIIrd Quadrant and (5, -1) lies in the IVth Quadrant.
Therefore, points (-4, 2),(1, 6), (-2, -4) and (5, —1) lie in different quadrants, making Assertion (A) as true.
All points with different values of x and y-coordinates may not lie in the different quadrants as for the points (2, 3) and (4, 1), respective values of their x and y-coordinates are different but these points lie in the first quadrant as they both are of the form (+a, +b). Therefore, Reason (R) is false.
Question 4.
Assertion (A): If an object is placed at a point (1, 2) and a mirror is placed along y-axis, then its image will be at the point (—1, 2).
Reason (R): The image of a point (a, b) in the mirror along x-axis is given by the point (a, -b).
Solution:
(b) Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).
Explanation:
Here, the image of a point (a, b) in the mirror along y-axis will be of the form (-a, b). So, the image of the object placed at the point (1, 2) in a mirror placed along y-axis will be at the point (-1, 2).
Therefore, Assertion (A) is true.
The image of a point (a, b) in the mirror along x-axis is given by the point (a, -b), is a fact. Therefore, Reason (R) is true.
But in this case Reason (R) is not the correct explanation of Assertion (A).
Question 5.
Assertion (A): If A(3, -3) and B(2, -3), then AB = 1 unit.
Reason (R): The distance between the points of the forms (u, v) and (w, v) is given by |w — u|.
Solution:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Explanation:
The distance between the points A(3, -3) and B(2, -3) is given by
AB = | 2 – 3 | = | —1| =1 unit. Therefore, Assertion (A) is true.
The distance between the points of the forms (u, v) and (w, v) is given by |w — u|. Therefore, Reason (R) is also true.
In this way 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, u) and (w, x) is given by
\(\sqrt{(w-u)^2+(x-v)^2}\) units.
Solution:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Explanation:
The distance between the points U(3, -7) and V(-2, -2) is given by UV =
\(\sqrt{(-2-3)^2+(-2+7)^2}=\sqrt{(-5))^2+(5)^2}=\sqrt{25+25}\)
= 5√2 units.
Therefore, Assertion (A) is true.
The distance between the points of the forms (;u, v) and (w, x) is given by V(ic – u)2+ (x + v)2 units.
Therefore, Reason (R) is true.
In this way Reason (R) is the correct explanation of Assertion (A).
Question 7.
Assertion (A): The distance of the point (-3, 5) from the x-axis is 3 units.
Reason (R): Ordinate of a point gives the distance of the point from the x-axis.
Solution:
(d) Assertion (A) is false but Reason (R) is true.
Explanation: The distance of the point (-3. 5) is 5 units from the x-axis.
Therefore, Assertion (A) is false.
Ordinate of a point gives the distance of the point from the x-axis is a fact.
Therefore, Reason (R) is true.
Question 8.
Assertion (A): For the point (1, -9), the value of (absecissa-ordinate) = 10.
Reason (R): In the coordinate of any point (x, y), x is called absecissa and y is called ordinate.
Solution:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Question 9.
Assertion (A): For 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.
Solution:
(b) Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).
Question 10.
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 IInd Quadrant.
Solution:
(d) Assertion (A) is false but Reason (R) is true.
Question 11.
Assertion (A): If an object is placed at a point (-1, 1) and a mirror is placed along 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 x-axis is given by the point (a, —b).
Solution:
(b) Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).
Question 12.
Assertion (A): If A(3, -3) and B(4, -3), then AB = 1 unit.
Reason (R): The distance between the points of the forms (u, v) and (w, v) is given by |w — u|.
Solution:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Question 13.
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.
Solution:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).