# CBSE Solutions for Class 10 Maths

#### Select CBSE Solutions for class 10 Subject & Chapters Wise :

If the distance between the points (2, –2) and (–1, x) is 5, one of the values of x is _________

2

The mid-point of the line segment joining the points A (–2, 8) and B (– 6, – 4) is ______

(– 4, 2)

The points A (9, 0), B (9, 6), C (–9, 6) and D (–9, 0) are the vertices of a ____________

Rectangle

The distance of the point P (2, 3) from the x-axis is______

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)

PQ = 10 …Given in question
PQ2 = 102 = 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)

AB = 10 units … [Given in the question
AB2 = 102 = 100 … [Squaring
(11 – 3)2 + (y + 1)2 = 100
82 + (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)

PA = QA …[Given in the question
PA2 = QA2 … [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)

Let P(2, 4), A(5, k) and B(k, 7).

PA = PB …[Given in the question
PA2 = PB2 … [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)

PA = PB …Given in the question
PA2 = PB2 … [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
k2 – 8k + 16 + 4 + k2 – 4k = 1 + 9
2k2 – 12k + 20 – 10 = 0
2k2 – 12k + 10 = 0
k2 – 6k + 5 = 0 …[Dividing by 2
k2 – 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

1-C, 2-D, 3-A, 4-B

Distance from origin

 1 A(5, −12) A 2√13 2 B(−5, 5) B 13 3 C(−4, −6) C 4 4 D(0,4) D 5√2

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

1-D, 2-A, 3-B, 4-C