Chapter 3:Pair of Linear Equations in Two Variables

Chapter 5: Arithmetic Progression

Chapter 8: Introduction to Trigonometry

Chapter 9:Some Applications of Trigonometry

Chapter 12:Area Related to Circles

Coordinate Geometry

(i) (2, 3), (4, 1)

(ii) (-5, 7), (-1, 3)

(iii) (a, b), (- a, – b)

Distance

.

.

=√(36-0)

=√36

=√ 1296 + 225

=√ 1521

=39 The distance between town A and B will be 39 km.

A = (1, 5) B = (2, 3) and C = (-2, -11)

. Hrer AB + BC ≠ CA

So the points (1, 5), (2, 3), and ( – 2, – 11) are not collinear.

Let the three point of triangle are A(5, – 2), B(6, 4), and C(7, – 2)

. Here AB = BC

here Two sides of isosceles triangle are equal .Given points are vertices of an isosceles triangle.

.

Using Distance formula

. AB = BC = CD = DA = 3√2

. ABCD is a square because all sides are of equal length so Champa was correct.

(i) (- 1, – 2), (1, 0), (- 1, 2), (- 3, 0)

(ii) (- 3, 5), (3, 1), (0, 3), (- 1, – 4)

(iii) (4, 5), (7, 6), (4, 3), (1, 2)

Side length = AB = BC = CD = DA = 2√2

Diagonal = AC = BD = 4

Therefore, the given points are the vertices of a square. (ii) Let the points (- 3, 5), (3, 1), (0, 3), and ( – 1, – 4) be representing the vertices A, B, C, and D of the given quadrilateral respectively.

All sides of this quadrilateral are of different then it is a general quadrilateral.

(iii) Let the points A(4, 5), B(7, 6), C(4, 3), and D(1, 2)

NCERT Solutions for Class 10 Chapter 7-9

Opposite sides of this quadrilateral are of the same length. and the diagonals are of different lengths. So the given points are the vertices of a parallelogram.

A(x, 0); B(2, – 5) and C(- 2, 9)

Therefore, the point is (- 7, 0).

Squaring both sides, 64 +(y+3)

(y+3)

y + 3 = ±6

y + 3 = +6 or y + 3 = −6

y = 3 or -9.

PQ = QR So

Squaring both the sides we,get

41 = x

xx

x = ± 4

x = 4 or x = -4

Coordinates of Point R(4, 6) OR

If R (4, 6), then QR

If R(-4, 6),then QR

We know in equidistant