If the distance between the points (2, –2) and (–1, x) is 5, one of the values of x is _________
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2
The mid-point of the line segment joining the points A (–2, 8) and B (– 6, – 4) is ______
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(– 4, 2)
The points A (9, 0), B (9, 6), C (–9, 6) and D (–9, 0) are the vertices of a ____________
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Rectangle
The distance of the point P (2, 3) from the x-axis is______
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3
Find that value(s) of x for which the distance between the points P(x, 4) and Q(9, 10) is 10 units. (2011D)
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PQ = 10 …Given in question
PQ^{2} = 10^{2} = 100 … [Squaring
(9 – x)^{2} + (10 – 4)^{2} = 100… (using the distance formula
(9 – x)^{2} + 36 = 100
(9 – x)^{2} = 100 – 36 = 64
(9 – x) = ± 8 …[Taking square-root
9 – x = 8 or 9 – x = -8
9 – 8 = x or 9+ 8 = x
x = 1 or x = 17
Find the value of y for which the distance between the points A (3,-1) and B (11, y) is 10 units. (2011OD)
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AB = 10 units … [Given in the question
AB^{2} = 10^{2} = 100 … [Squaring
(11 – 3)^{2} + (y + 1)^{2} = 100
8^{2} + (y + 1)^{2} = 100
(y + 1)^{2} = 100 – 64 = 36
y + 1 = ±6 … [Taking square-root
y = -1 ± 6 ∴ y = -7 or 5
The point A(3, y) is equidistant from the points P(6, 5) and Q(0, -3). Find the value of y. (2011D)
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PA = QA …[Given in the question
PA^{2} = QA^{2} … [Squaring
(3 – 6)^{2} + (y – 5)^{2} = (3 – 0)^{2} + (y + 3)^{2}
9 + (y – 5)^{2} = 9 + (y + 3)^{2}
(y – 5)^{2} = (y + 3)^{2}
y – 5 = ±(y + 3) … [Taking square root
y – 5 = y + 3 y – 5 = -y – 3
0 = 8 … which is not possible ∴ y = 1
Find the value of k, if the point P(2, 4) is equidistant from the points A(5, k) and B(k, 7). (2012OD)
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Let P(2, 4), A(5, k) and B(k, 7).
PA = PB …[Given in the question
PA^{2} = PB^{2} … [Squaring
(5 – 2)^{2} + (k – 4)^{2} = (k – 2)^{2} + (7 – 4)^{2}
9 + (k – 4)^{2} – (k – 2)^{2} = 9
(k – 4 + k – 2) (k – 4 – k + 2) = 0
(2k – 6)(-2) = 0
2k – 6 = 0
2k = 6 ∴ k = 3
If the point P(k – 1, 2) is equidistant from the points A(3, k) and B(k, 5), find the values of k. (2014OD)
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PA = PB …Given in the question
PA^{2} = PB^{2} … [Squaring
⇒ (k – 1 – 3)^{2} + (2 – k)^{2} = (k – 1 – k)^{2} + (2 – 5)^{2}
⇒ (k – 4)^{2} + (2 – k)^{2} = (-1)^{2} + (-3)^{2}
k^{2} – 8k + 16 + 4 + k^{2} – 4k = 1 + 9
2k^{2} – 12k + 20 – 10 = 0
2k^{2} – 12k + 10 = 0
⇒ k^{2} – 6k + 5 = 0 …[Dividing by 2
⇒ k^{2} – 5k – k + 5 = 0
⇒ k(k – 5) – 1(k – 5) = 0
⇒ (k – 5) (k – 1) = 0
⇒ k – 5 = 0 or k – 1 = 0
∴ k = 5 or k = 1
Match the distance
1 |
A(9, 3) and B(15, 11) |
A |
17 |
2 |
A(7, −4) and B(−5, 1) |
B |
3√2 |
3 |
A(−6, −4) and B(9, −12) |
C |
10 |
4 |
A(1, −3) and B(4, −6) |
D |
13 |
Answer :
1-C, 2-D, 3-A, 4-B
Answer :
1-B, 2-D, 3-A, 4-C
Distance between the points
1 |
A(2, −1) and B(5, 3) |
A |
10 |
2 |
A(2,−3) and B(10,-9) |
B |
√10 |
3 |
A(0, 2) and B(3, 1) |
C |
2√13 |
4 |
A(6, 5) and B(0,9) |
D |
5 |
Answer :
1-D, 2-A, 3-B, 4-C
This chapter deals with finding the area between two points whose coordinate values are provided. For instance the area of a triangle. This chapter has some basic concepts like the area of a triangle, rhombus, the distance between sides, and intersections. This chapter teaches you the relationship between numerical and geometry and their application in our daily lives.
The GSEB Books for class 10 are designed as per the syllabus followed Gujarat Secondary and Higher Secondary Education Board provides key detailed, and a through solutions to all the questions relating to the GSEB textbooks.
The purpose is to provide help to the students with their homework, preparing for the examinations and personal learning. These books are very helpful for the preparation of examination.
For more details about the GSEB books for Class 10, you can access the PDF which is as in the above given links for the same.