{"id":29255,"date":"2024-02-27T10:28:51","date_gmt":"2024-02-27T04:58:51","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=29255"},"modified":"2024-02-28T09:41:15","modified_gmt":"2024-02-28T04:11:15","slug":"ap-inter-1st-year-maths-1b-solutions-chapter-2-ex-2a-in-telugu","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/ap-inter-1st-year-maths-1b-solutions-chapter-2-ex-2a-in-telugu\/","title":{"rendered":"AP Inter 1st Year Maths 1B Solutions Chapter 2 \u0c05\u0c15\u0c4d\u0c37 \u0c2a\u0c30\u0c3f\u0c35\u0c30\u0c4d\u0c24\u0c28\u0c02 Ex 2(a)"},"content":{"rendered":"

Practicing the\u00a0Intermediate 1st Year Maths 1B Textbook Solutions<\/a> Chapter 2 \u0c05\u0c15\u0c4d\u0c37 \u0c2a\u0c30\u0c3f\u0c35\u0c30\u0c4d\u0c24\u0c28\u0c02 Exercise 2(a) will help students to clear their doubts quickly.<\/p>\n

AP Inter 1st Year Maths 1B Solutions Chapter 2 \u0c05\u0c15\u0c4d\u0c37 \u0c2a\u0c30\u0c3f\u0c35\u0c30\u0c4d\u0c24\u0c28\u0c02 Exercise 2(a)<\/h2>\n

\u0c05\u0c2d\u0c4d\u0c2f\u0c3e\u0c38\u0c02 \u2013 2 (\u0c0e)<\/span><\/p>\n

I.<\/span><\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 1.
\n\u0c05\u0c15\u0c4d\u0c37\u0c3e\u0c32 \u0c38\u0c2e\u0c3e\u0c02\u0c24\u0c30\u0c2a\u0c30\u0c3f\u0c35\u0c30\u0c4d\u0c24\u0c28 \u0c26\u0c4d\u0c35\u0c3e\u0c30\u0c3e \u0c2e\u0c42\u0c32\u0c2c\u0c3f\u0c02\u0c26\u0c41\u0c35\u0c41\u0c28\u0c41 (4, \u22125) \u0c15\u0c41 \u0c2e\u0c3e\u0c30\u0c3f\u0c38\u0c4d\u0c24\u0c47 \u0c15\u0c4a\u0c24\u0c4d\u0c24 \u0c05\u0c15\u0c4d\u0c37\u0c3e\u0c32 \u0c26\u0c43\u0c37\u0c4d\u0c1f\u0c4d\u0c2f\u0c3e \u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c2c\u0c3f\u0c02\u0c26\u0c41\u0c35\u0c41\u0c32 \u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c15\u0c3e\u0c32\u0c41 \u0c15\u0c28\u0c41\u0c15\u0c4d\u0c15\u0c4b\u0c02\u0c21\u0c3f.
\ni) (0, 3),
\nii) (\u22122, 4)
\niii) (4, -5)
\n\u0c38\u0c3e\u0c27\u0c28:
\ni) \u0c15\u0c4a\u0c24\u0c4d\u0c24 \u0c2e\u0c42\u0c32\u0c2c\u0c3f\u0c02\u0c26\u0c41\u0c35\u0c41 = (4, -5); h = 4, k = -5
\n\u0c2a\u0c3e\u0c24 \u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c15\u0c3e\u0c32\u0c41 (0, 3)
\n\u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c15\u0c3e\u0c15\u0c4d\u0c37\u0c3e\u0c32\u0c15\u0c41 8 \u0c15\u0c4b\u0c23\u0c02 \u0c2d\u0c4d\u0c30\u0c2e\u0c23\u0c02 \u0c1a\u0c47\u0c2f\u0c35\u0c32\u0c46\u0c28\u0c41.
\ntan 2\u03b8 = \\(\\frac{2 h}{a-b}\\) ; x = 0, y = 3
\nx’ = x – h= 0 – 4 = -4
\ny’ = y – k = 3 + 5 = 8
\n\u0c28\u0c42\u0c24\u0c28 \u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c15\u0c3e\u0c32\u0c41 (-4, 8)<\/p>\n

ii) \u0c2a\u0c3e\u0c24 \u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c15\u0c3e\u0c32\u0c41 (-2, 4)
\nx = -2, y = 4
\nx’ = x – h = -2 -4 = -6
\ny’ = y – k = 4 + 5 = 9
\n\u0c28\u0c42\u0c24\u0c28 \u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c15\u0c3e\u0c32\u0c41 (-6, 9)<\/p>\n

iii) \u0c2a\u0c3e\u0c24 \u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c15\u0c3e\u0c32\u0c41 (4, -5)
\nx = 4, y = -5
\nx’ =x-h=4 – 4 = 0
\ny’ = y – k = -5 + 5 = 0
\n\u0c28\u0c42\u0c24\u0c28 \u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c15\u0c3e\u0c32\u0c41 (0, 0)<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 2.
\n\u0c05\u0c15\u0c4d\u0c37\u0c3e\u0c32 \u0c38\u0c2e\u0c3e\u0c02\u0c24\u0c30 \u0c2a\u0c30\u0c3f\u0c35\u0c30\u0c4d\u0c24\u0c28\u0c26\u0c4d\u0c35\u0c3e\u0c30\u0c3e \u0c2e\u0c42\u0c32\u0c2c\u0c3f\u0c02\u0c26\u0c41\u0c35\u0c41 (2, 3) \u0c15\u0c41 \u0c2e\u0c3e\u0c30\u0c3f\u0c02\u0c26\u0c3f. P \u0c2c\u0c3f\u0c02\u0c26\u0c41\u0c35\u0c41 \u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c15\u0c3e\u0c32\u0c41 \u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3f\u0c27\u0c02\u0c17\u0c3e \u0c2e\u0c3e\u0c30\u0c3f\u0c24\u0c47, \u0c2e\u0c42\u0c32 \u0c35\u0c4d\u0c2f\u0c35\u0c38\u0c4d\u0c25\u0c32\u0c4b P \u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c15\u0c3e\u0c32\u0c41 \u0c15\u0c28\u0c41\u0c15\u0c4d\u0c15\u0c4b\u0c02\u0c21\u0c3f.
\ni) (4, 5)
\nii) (4, -3),
\niii) (0, 0)
\n\u0c38\u0c3e\u0c27\u0c28:
\ni) \u0c28\u0c42\u0c24\u0c28 \u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c15\u0c3e\u0c32\u0c41 (4, 5)
\nx’ = 4, y’ = 5
\nx = x + h = 4 + 2= 6
\ny = y’ + k = 5 + 3 = 8
\n\u0c2a\u0c3e\u0c24 \u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c15\u0c3e\u0c32\u0c41 (6, 8)<\/p>\n

ii) \u0c28\u0c42\u0c24\u0c28 \u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c15\u0c3e\u0c32\u0c41 (-4, 3)
\nx’ = -4, y’ = 3
\nx = x’ + h = 4 + 2 = -2
\ny = y’ + k = 3 + 3 = 6
\n\u0c2a\u0c3e\u0c24 \u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c15\u0c3e\u0c32\u0c41 (\u22122, 6)<\/p>\n

iii) \u0c28\u0c42\u0c24\u0c28 \u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c15\u0c3e\u0c32\u0c41 (0, 0)
\nx’ = 0, y’ = 0
\nx = x’ + h = 0 + 2 = 2
\ny = y’ + k = 0 + 3 = 3
\n\u0c2a\u0c3e\u0c24 \u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c15\u0c3e\u0c32\u0c41 (2, 3)<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 3.
\n\u0c05\u0c15\u0c4d\u0c37\u0c3e\u0c32 \u0c38\u0c2e\u0c3e\u0c02\u0c24\u0c30 \u0c2a\u0c30\u0c3f\u0c35\u0c30\u0c4d\u0c24\u0c28 \u0c26\u0c4d\u0c35\u0c3e\u0c30\u0c3e \u0c2c\u0c3f\u0c02\u0c26\u0c41\u0c35\u0c41 (3, 0)\u0c28\u0c41 (2, \u22123) \u0c15\u0c41 \u0c2e\u0c3e\u0c30\u0c4d\u0c1a\u0c21\u0c3e\u0c28\u0c3f\u0c15\u0c3f \u0c2e\u0c42\u0c32\u0c2c\u0c3f\u0c02\u0c26\u0c41\u0c35\u0c41\u0c28\u0c41 \u0c0f\u0c2c\u0c3f\u0c02\u0c26\u0c41\u0c35\u0c41\u0c15\u0c41 \u0c2e\u0c3e\u0c30\u0c4d\u0c1a\u0c3e\u0c32\u0c4b \u0c24\u0c46\u0c32\u0c2a\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28:
\n(x, y) = (3, 0)
\n(x’, y’) = (2, \u22123)
\n\u0c2e\u0c42\u0c32 \u0c2c\u0c3f\u0c02\u0c26\u0c41\u0c35\u0c41 (h, k) \u0c15\u0c41 \u0c2e\u0c3e\u0c30\u0c4d\u0c1a\u0c35\u0c32\u0c46\u0c28\u0c41
\nh = x – x’ = 3 – 2 = 1
\nk = y – y = 0 + 3 = 3
\n\u2234 (h, k) = (1, 3)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 4.
\n\u0c05\u0c15\u0c4d\u0c37\u0c3e\u0c32 \u0c38\u0c2e\u0c3e\u0c02\u0c24\u0c30 \u0c2a\u0c30\u0c3f\u0c35\u0c30\u0c4d\u0c24\u0c28 \u0c26\u0c4d\u0c35\u0c3e\u0c30\u0c3e \u0c2e\u0c42\u0c32\u0c2c\u0c3f\u0c02\u0c26\u0c41\u0c35\u0c41\u0c28\u0c41 (\u22121, 2) \u0c15\u0c41 \u0c2e\u0c3e\u0c30\u0c3f\u0c38\u0c4d\u0c24\u0c47 \u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c32 \u0c30\u0c42\u0c2a\u0c3e\u0c02\u0c24\u0c30\u0c3e\u0c32\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c15\u0c4d\u0c15\u0c4b\u0c02\u0c21\u0c3f.
\ni) x2<\/sup> + y2<\/sup> + 2x – 4y + 1 = 0
\nii) 2x2<\/sup> + y2<\/sup> – 4x + 4y = 0
\n\u0c38\u0c3e\u0c27\u0c28:
\ni) \u0c26\u0c24\u0c4d\u0c24 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c2e\u0c41
\nx2<\/sup> + y2<\/sup> + 2x – 4y + 1 = 0
\n\u0c2e\u0c42\u0c32 \u0c2c\u0c3f\u0c02\u0c26\u0c41\u0c35\u0c41 (\u22121, 2) \u0c15\u0c41 \u0c2e\u0c3e\u0c30\u0c4d\u0c1a\u0c35\u0c32\u0c46\u0c28\u0c41
\nh = -1, k = 2
\n\u0c2a\u0c30\u0c3f\u0c35\u0c30\u0c4d\u0c24\u0c28 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c3e\u0c32\u0c41
\nx = x’ + h, y = y + k
\ni.e., x = x’ – 1, y = y + 2
\n\u0c30\u0c42\u0c2a\u0c3e\u0c02\u0c24\u0c30 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c2e\u0c41 (x’ – 1)2<\/sup> + (y + 2)2<\/sup>
\n+ 2(x’ \u2013 1) \u2013 4(y’ + 2) + 1 = 0
\n(x’)2<\/sup> + 1 \u2212 2x’ + (y’)2<\/sup> + 4 + 4y’ + 2x’ – 2 – 4y’ – 8 + 1 = 0
\n(x’)2<\/sup> + (y’)2<\/sup> \u2212 4 = 0<\/p>\n

ii) \u0c2a\u0c3e\u0c24 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c2e\u0c41 2x2<\/sup> + y2<\/sup> – 4x + 4y = 0
\n\u0c28\u0c42\u0c24\u0c28 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c2e\u0c41 2 (x’ – 1)2<\/sup> + (y + 2)2<\/sup> -4(x’ – 1) + 4(y + 2) = 0
\n2[(x’)2<\/sup> + 1 – 2x’] + (y’)2<\/sup> + 4 + 4y’ – 4x’ + 4 + 4y’ + 8 = 0
\n2(x’)2<\/sup> + 2 – 4x’ + (y’) 2<\/sup> + 4 + 4y’ – 4x + 4 + 4y’ + 8 = 0
\n2(x’)2<\/sup> + (y’)2<\/sup> = 8x’ + 8y + 18 = 0<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 5.
\n\u0c05\u0c15\u0c4d\u0c37\u0c3e\u0c32 \u0c38\u0c2e\u0c3e\u0c02\u0c24\u0c30 \u0c2a\u0c30\u0c3f\u0c35\u0c30\u0c4d\u0c24\u0c28 \u0c26\u0c4d\u0c35\u0c3e\u0c30\u0c3e \u0c2e\u0c42\u0c32\u0c2c\u0c3f\u0c02\u0c26\u0c41\u0c35\u0c41\u0c28\u0c41 \u0c0f \u0c2c\u0c3f\u0c02\u0c26\u0c41\u0c35\u0c41\u0c15\u0c41 \u0c2e\u0c3e\u0c30\u0c4d\u0c1a\u0c3f\u0c02\u0c26\u0c40, \u0c24\u0c26\u0c4d\u0c35\u0c3e\u0c30\u0c3e \u0c30\u0c42\u0c2a\u0c3e\u0c02\u0c24\u0c30\u0c02 \u0c1a\u0c46\u0c02\u0c26\u0c3f\u0c28 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c2e\u0c41 \u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26 \u0c07\u0c35\u0c4d\u0c35\u0c21\u0c2e\u0c48\u0c02\u0c26\u0c3f. \u0c2e\u0c42\u0c32\u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c3e\u0c28\u0c4d\u0c28\u0c3f \u0c15\u0c28\u0c41\u0c15\u0c4d\u0c15\u0c4b\u0c02\u0c21\u0c3f.
\ni) (3, – 4); x2<\/sup> + y2<\/sup> = 4
\nii) (-1, 2); x2<\/sup> + 2 y2<\/sup> + 16 = 0.
\n\u0c38\u0c3e\u0c27\u0c28:
\ni) \u0c2e\u0c42\u0c32 \u0c2c\u0c3f\u0c02\u0c26\u0c41\u0c35\u0c41\u0c28\u0c41 = (3, -4) = (h, k) \u0c15\u0c41 \u0c2e\u0c3e\u0c30\u0c4d\u0c1a\u0c35\u0c32\u0c46\u0c28\u0c41
\nx’ = x -h,
\n= x – 3
\ny’ = y-k
\n= y + 4
\n\u0c2e\u0c42\u0c32 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c2e\u0c41 (x’)2<\/sup> + (y’)2<\/sup> = 4
\n(x – 3)2<\/sup> + (y + 4) 2<\/sup> = 4
\nx2<\/sup> – 6x + 9 + y2<\/sup> + 8y + 16
\n\u2234 x2<\/sup> + y2<\/sup> – 6x + 8y + 21 = 0<\/p>\n

ii) \u0c2e\u0c42\u0c32 \u0c2c\u0c3f\u0c02\u0c26\u0c41\u0c35\u0c41 = (h, k) = (-1, 2) \u0c15\u0c41 \u0c2e\u0c3e\u0c30\u0c4d\u0c1a\u0c35\u0c32\u0c46\u0c28\u0c41.
\nx’ = x – h,
\n= x + 1
\ny’ = y – k
\n= y – 2
\n= 4
\nx’2<\/sup> + 2y’2<\/sup> + 16 = 0 \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c2e\u0c42\u0c32 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c2e\u0c41
\n(x + 1)2<\/sup> + 2(y – 2)2<\/sup> + 16 = 0
\nx2<\/sup> + 2x + 1 + 2y2<\/sup> – 8y + 8 + 16 = 0
\nx2<\/sup> + 2y2<\/sup> + 2x – 8y + 25 = 0<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 6.
\n4x2<\/sup> + 9y2<\/sup> – 8x + 36y + 4 = 0 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c02\u0c32\u0c4b \u0c2e\u0c4a\u0c26\u0c1f\u0c3f \u0c24\u0c30\u0c17\u0c24\u0c3f \u0c2a\u0c26\u0c3e\u0c32\u0c41 \u0c32\u0c4b\u0c2a\u0c3f\u0c02\u0c2a\u0c1a\u0c47\u0c2f\u0c21\u0c3e\u0c28\u0c3f\u0c15\u0c3f \u0c2e\u0c42\u0c32 \u0c2c\u0c3f\u0c02\u0c26\u0c41\u0c35\u0c41\u0c28\u0c41 \u0c0f \u0c2c\u0c3f\u0c02\u0c26\u0c41\u0c35\u0c41\u0c15\u0c41 \u0c2e\u0c3e\u0c30\u0c4d\u0c1a\u0c3e\u0c32\u0c4b \u0c15\u0c28\u0c41\u0c15\u0c4d\u0c15\u0c4b\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28:
\n\u0c26\u0c24\u0c4d\u0c24 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c2e\u0c41
\n4x2<\/sup> + 9y2<\/sup> – 8x + 36y + 4 = 0
\na = 4
\nb = 9
\ng = – 4
\nf = 18
\n– \\(\\frac{g}{a}=\\frac{4}{4}\\) = 1, –\\(\\frac{f}{b}=-\\frac{18}{9}\\) = -2
\n\u0c2e\u0c42\u0c32 \u0c2c\u0c3f\u0c02\u0c26\u0c41\u0c35\u0c41 (1, 2) \u0c15\u0c41 \u0c2a\u0c30\u0c3f\u0c35\u0c30\u0c4d\u0c24\u0c28 \u0c1a\u0c47\u0c2f\u0c35\u0c32\u0c46\u0c28\u0c41.<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 7.
\n30\u00b0 \u0c15\u0c4b\u0c23\u0c02\u0c24\u0c4b \u0c05\u0c15\u0c4d\u0c37\u0c3e\u0c32\u0c28\u0c41 \u0c2d\u0c4d\u0c30\u0c2e\u0c23\u0c02 \u0c1a\u0c47\u0c38\u0c3f\u0c28\u0c2a\u0c4d\u0c2a\u0c41\u0c21\u0c41, \u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c2c\u0c3f\u0c02\u0c26\u0c41\u0c35\u0c41\u0c32 \u0c15\u0c4a\u0c24\u0c4d\u0c24 \u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c15\u0c3e\u0c32\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c15\u0c4d\u0c15\u0c4b\u0c02\u0c21\u0c3f.
\ni) (0, 5)
\nii) (\u22122, 4)
\niii) (0, 0)
\n\u0c38\u0c3e\u0c27\u0c28:
\ni) \u03b8 = 30\u00b0 \u0c05\u0c28\u0c3f \u0c2f\u0c3f\u0c35\u0c4d\u0c35\u0c2c\u0c21\u0c3f\u0c02\u0c26\u0c3f. \u0c2a\u0c3e\u0c24 \u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c15\u0c3e\u0c32\u0c41 (0, 5)
\nx = 0, y = 5
\nx’ = x. cos \u03b8 + y. sin \u03b8
\n= 0. cos 30\u00b0 + 5. sin 30\u00b0 = \\(\\frac{5}{2}\\)
\ny’ = -x sin \u03b8 + y cos \u03b8
\n=-0. sin 30\u00b0 + 5 cos 30\u00b0 = \\(\\frac{5 \\sqrt{3}}{2}\\)
\n\u0c2a\u0c30\u0c3f\u0c35\u0c30\u0c4d\u0c24\u0c28 \u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c15\u0c3e\u0c32\u0c41 \\(\\left(\\frac{5}{2}, \\frac{5 \\sqrt{3}}{2}\\right)\\)<\/p>\n

ii) \u0c2a\u0c3e\u0c24 \u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c15\u0c3e\u0c32\u0c41 (-2, 4)
\nx = -2, y = 4
\nx’ = x cos \u03b8 + y sin \u03b8
\n= (-2). cos 30\u00b0 + 4. sin 30\u00b0
\n= -2 . \\(\\frac{\\sqrt{3}}{2}\\) + 4 . \\(\\frac{1}{2}\\) = –\\(\\sqrt{3}\\) + 2
\ny’ = – x sin \u03b8 + y cos \u03b8
\n= (\u2212 2). sin 30\u00b0 + 4. cos 30\u00b0
\n= -2 . \\(\\frac{1}{2}\\) + 4 . \\(\\frac{\\sqrt{3}}{2}\\)
\n= 1 + 2\\(\\sqrt{3}\\)
\n\u2234 \u0c2a\u0c30\u0c3f\u0c35\u0c30\u0c4d\u0c24\u0c28 \u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c15\u0c3e\u0c32\u0c41 (-\\(\\sqrt{3}\\) + 2, 1+ 2\\(\\sqrt{3}\\))<\/p>\n

iii) (x, y) = (0, 0) \u0c2e\u0c30\u0c3f\u0c2f\u0c41 \u03b8 = 30\u00b0 \u0c05\u0c28\u0c3f \u0c2f\u0c3f\u0c35\u0c4d\u0c35\u0c2c\u0c21\u0c3f\u0c02\u0c26\u0c3f.
\nx = (0, y) \u21d2 x = x. cos 30\u00b0 – y sin 30\u00b0
\n= 0 . \\(\\frac{\\sqrt{3}}{2}\\) – 0 . \\(\\frac{1}{2}\\) = 0
\ny = x. sin 30\u00b0 + y.cos 30\u00b0
\n= 0 . \\(\\frac{1}{2}\\) + 0 . \\(\\frac{\\sqrt{3}}{2}\\) = 0
\n\u0c2a\u0c30\u0c3f\u0c35\u0c30\u0c4d\u0c24\u0c28 \u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c15\u0c3e\u0c32\u0c41 (0, 0)<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 8.
\n60\u00b0 \u0c15\u0c4b\u0c23\u0c02\u0c24\u0c4b \u0c05\u0c15\u0c4d\u0c37\u0c3e\u0c32\u0c28\u0c41 \u0c2d\u0c4d\u0c30\u0c2e\u0c23\u0c02 \u0c1a\u0c47\u0c38\u0c3f\u0c28\u0c2a\u0c4d\u0c2a\u0c41\u0c21\u0c41, \u0c2e\u0c42\u0c32 \u0c2c\u0c3f\u0c02\u0c26\u0c41\u0c35\u0c41\u0c32 \u0c15\u0c4a\u0c24\u0c4d\u0c24 \u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c15\u0c3e\u0c32\u0c28\u0c41 \u0c07\u0c35\u0c4d\u0c35\u0c21\u0c2e\u0c48\u0c02\u0c26\u0c3f.
\ni) (3, 4),
\nii) (-7, 2)
\niii) (2, 0) \u0c2e\u0c42\u0c32 \u0c35\u0c4d\u0c2f\u0c35\u0c38\u0c4d\u0c25\u0c32\u0c4b \u0c08 \u0c2c\u0c3f\u0c02\u0c26\u0c41\u0c35\u0c41\u0c32 \u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c15\u0c3e\u0c32\u0c41 \u0c15\u0c28\u0c41\u0c15\u0c4d\u0c15\u0c4b\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28:
\ni) \u03b8 = 60\u00b0 \u0c05\u0c28\u0c3f \u0c2f\u0c3f\u0c35\u0c4d\u0c35\u0c2c\u0c21\u0c3f\u0c02\u0c26\u0c3f.
\n\u0c28\u0c42\u0c24\u0c28 \u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c15\u0c3e\u0c32\u0c41 (3, 4)
\nx’ = 3, y’ = 4
\nx = x’ cos \u03b8 – y’ sin \u03b8
\n= 3. cos 60\u00b0 – 4. sin 60\u00b0
\n= 3 . \\(\\frac{1}{2}\\) – \\(\\frac{4 \\cdot \\sqrt{3}}{2}\\) = \\(\\frac{3-4 \\sqrt{3}}{2}\\)<\/p>\n

y = x’ sin \u03b8 + y’ cos \u03b8
\n= 3 sin 60\u00b0 + 4. cos 60\u00b0
\n= \\(\\text { 3. } \\frac{\\sqrt{3}}{2}+4 \\cdot \\frac{1}{2}=\\frac{4+3 \\sqrt{3}}{2}\\)
\nP \u0c24\u0c4a\u0c32\u0c3f \u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c15\u0c3e\u0c32\u0c41 \\(\\left(\\frac{3-4 \\sqrt{3}}{2}, \\frac{4+3 \\sqrt{3}}{2}\\right)\\)<\/p>\n

ii) \u0c28\u0c42\u0c24\u0c28 \u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c15\u0c3e\u0c32\u0c41 (\u22127, 2)
\nx’ = 7, y’ = 2
\nx = x’ cos \u03b8 – y’ sin \u03b8
\n= (-7) cos 60\u00b0 – 2. sin 60\u00b0
\n= -7 . \\(\\frac{1}{2}-2 \\cdot \\frac{\\sqrt{3}}{2}=\\frac{-7-2 \\sqrt{3}}{2}\\)
\ny = x’ sin \u03b8 + y’. cos \u03b8
\n= -7. sin 60\u00b0 + 2. cos 60\u00b0
\n= -7 . \\(\\frac{\\sqrt{3}}{2}+2 \\cdot \\frac{1}{2}=\\frac{2-7 \\sqrt{3}}{2}\\)
\nQ \u0c24\u0c4a\u0c32\u0c3f \u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c15\u0c3e\u0c32\u0c41 \\(\\left(\\frac{-7-2 \\sqrt{3}}{2}, \\frac{2-7 \\sqrt{3}}{2}\\right)\\)<\/p>\n

iii) \u0c28\u0c42\u0c24\u0c28 \u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c15\u0c3e\u0c32\u0c41 (2, 0)
\nx’ = 2, y’ = 0
\nx = x’ cos \u03b8 – y’ sin \u03b8
\n= 2. cos 60\u00b0 – 0. sin 60\u00b0
\n= 2 . \\(\\frac{1}{2}\\) – 0 . \\(\\frac{\\sqrt{3}}{2}\\) = 1 – 0 = 1
\ny = x’ sin \u03b8 + y’ cos \u03b8
\n= 2. sin 60\u00b0 + 0. cos 60\u00b0
\n= 2. \\(\\frac{\\sqrt{3}}{2}\\) + 0 . \\(\\frac{1}{2}\\) = \\(\\sqrt{3}\\)
\nQ \u0c24\u0c4a\u0c32\u0c3f \u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c15\u0c3e\u0c32\u0c41 (1, \\(\\sqrt{3}\\))<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 9.
\nx2<\/sup> + 4xy + y2<\/sup> – 2x + 2y – 60 xy \u0c2a\u0c26\u0c02 \u0c32\u0c4b\u0c2a\u0c3f\u0c02\u0c2a\u0c1a\u0c47\u0c2f\u0c21\u0c3e\u0c28\u0c3f\u0c15\u0c3f \u0c05\u0c15\u0c4d\u0c37\u0c3e\u0c32\u0c28\u0c41 \u0c0f \u0c15\u0c4b\u0c23\u0c02\u0c32\u0c4b \u0c2d\u0c4d\u0c30\u0c2e\u0c23 \u0c2a\u0c30\u0c3f\u0c35\u0c30\u0c4d\u0c24\u0c28 \u0c1a\u0c47\u0c2f\u0c3e\u0c32\u0c4b \u0c15\u0c28\u0c41\u0c15\u0c4d\u0c15\u0c4b\u0c02\u0c21\u0c3f. [June ’04]
\n\u0c38\u0c3e\u0c27\u0c28:
\nx2<\/sup> + 4xy + y2<\/sup> – 2x + 2y – 6 = 0
\nax2<\/sup> + 2hxy + by2<\/sup> + 2gx + 2fy + c = 0 \u0c24\u0c4b
\n\u0c2a\u0c4b\u0c32\u0c4d\u0c1a\u0c17\u0c3e a = 1, h = 2, b = 1, g = 1, f = 1, c = -6
\n\u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c15\u0c3e\u0c15\u0c4d\u0c37\u0c3e\u0c32\u0c15\u0c41 ‘\u03b8’ \u0c15\u0c4b\u0c23\u0c02 \u0c2d\u0c4d\u0c30\u0c2e\u0c23\u0c02 \u0c1a\u0c47\u0c38\u0c4d\u0c24\u0c47
\n\"AP<\/p>\n

II.<\/span><\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 1.
\n\u0c05\u0c15\u0c4d\u0c37\u0c3e\u0c32 \u0c38\u0c2e\u0c3e\u0c02\u0c24\u0c30 \u0c2a\u0c30\u0c3f\u0c35\u0c30\u0c4d\u0c24\u0c28 \u0c26\u0c4d\u0c35\u0c3e\u0c30\u0c3e \u0c2e\u0c42\u0c32 \u0c2c\u0c3f\u0c02\u0c26\u0c41\u0c35\u0c41\u0c28\u0c41 (2, 3) \u0c15\u0c41 \u0c2e\u0c3e\u0c30\u0c4d\u0c1a\u0c3f\u0c28\u0c2a\u0c4d\u0c2a\u0c41\u0c21\u0c41, \u0c12\u0c15 \u0c35\u0c15\u0c4d\u0c30\u0c02 \u0c30\u0c42\u0c2a\u0c3e\u0c02\u0c24\u0c30\u0c02 \u0c1a\u0c46\u0c02\u0c26\u0c3f\u0c28 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c2e\u0c41 x2<\/sup> + 3xy – 2y2<\/sup> + 17x – 7y – 11 = 0 \u0c05\u0c2f\u0c3f\u0c24\u0c47 \u0c35\u0c15\u0c4d\u0c30\u0c02 \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c2e\u0c42\u0c32 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c2e\u0c41 \u0c15\u0c28\u0c41\u0c15\u0c4d\u0c15\u0c4b\u0c02\u0c21\u0c3f. [May ’11]
\n\u0c38\u0c3e\u0c27\u0c28:
\n\u0c2a\u0c30\u0c3f\u0c35\u0c30\u0c4d\u0c24\u0c28 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c3e\u0c32\u0c41
\nx = x + h, y = y + k
\nx’ = x – h = x – 2, y’ = y – 3
\n\u0c2a\u0c30\u0c3f\u0c35\u0c30\u0c4d\u0c24\u0c3f\u0c24 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c2e\u0c41
\nx2<\/sup> + 3xy – 2y2<\/sup> + 17x – 7y – 11 = 0
\n\u0c24\u0c4a\u0c32\u0c3f \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c2e\u0c41
\n(x – 2)2<\/sup> + 3(x – 2) (y – 3) – 2(y – 3)2<\/sup> + 17(x – 2) – 7(y – 3) – 11 = 0
\nx2<\/sup> – 4x + 4 + 3xy – 9x – 6y + 18 – 2y2<\/sup> + 12y – 18 + 17x – 34 – 7y + 21 \u2212 11 = 0
\nx2<\/sup> + 3xy – 2y2<\/sup> + 4x – y – 20 = 0
\n\u0c07\u0c26\u0c3f \u0c15\u0c3e\u0c35\u0c32\u0c38\u0c3f\u0c28 \u0c2e\u0c42\u0c32 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c2e\u0c41<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 2.
\n45\u00b0\u0c15\u0c4b\u0c23\u0c02\u0c24\u0c4b \u0c05\u0c15\u0c4d\u0c37\u0c3e\u0c32\u0c28\u0c41 \u0c2d\u0c4d\u0c30\u0c2e\u0c23\u0c02 \u0c1a\u0c47\u0c38\u0c3f\u0c28\u0c2a\u0c4d\u0c2a\u0c41\u0c21\u0c41, \u0c30\u0c42\u0c2a\u0c3e\u0c02\u0c24\u0c30\u0c02 \u0c1a\u0c46\u0c02\u0c26\u0c3f\u0c28 \u0c35\u0c15\u0c4d\u0c30\u0c02 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c2e\u0c41 17x2<\/sup> – 16xy + 17y = 225. \u0c35\u0c15\u0c4d\u0c30\u0c02 \u0c2e\u0c42\u0c32 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c3e\u0c28\u0c4d\u0c28\u0c3f \u0c15\u0c28\u0c41\u0c15\u0c4d\u0c15\u0c4b\u0c02\u0c21\u0c3f. [T.S Mar. ’15; May ’12]
\n\u0c38\u0c3e\u0c27\u0c28:
\n\u0c2d\u0c4d\u0c30\u0c2e\u0c23 \u0c15\u0c4b\u0c23\u0c2e\u0c41 = \u03b8 = 45
\nx’ = x cos \u03b8 + y sin \u03b8 = x cos 45 + y sin 45
\n= \\(\\frac{x+y}{\\sqrt{2}}\\)
\ny’ = -x sin \u03b8 + y cos \u03b8 = \u2212 -x sin 45 + y cos 45
\n= \\(\\frac{-x+y}{\\sqrt{2}}\\)
\n17x2<\/sup> – 16xy + 17y2<\/sup> = 225 \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c24\u0c4a\u0c32\u0c3f \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c2e\u0c41
\n\"AP
\n17x2<\/sup> + 17y2<\/sup> + 34xy – 16y2<\/sup> + 16x2<\/sup> + 17x2<\/sup> + 17y2<\/sup> – 34xy = 450
\n50x2<\/sup> + 18y2<\/sup> = 450
\n\u0c24\u0c4a\u0c32\u0c3f \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c2e\u0c41
\n25x2<\/sup> + 9y2<\/sup> = 225<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 3.
\n\u0c05\u0c15\u0c4d\u0c37\u0c3e\u0c32\u0c28\u0c41 a \u0c15\u0c4b\u0c23\u0c02\u0c24\u0c4b \u0c2d\u0c4d\u0c30\u0c2e\u0c23\u0c02 \u0c1a\u0c47\u0c38\u0c3f\u0c28\u0c2a\u0c4d\u0c2a\u0c41\u0c21\u0c41, x cos \u03b1. + y sin \u03b1 = p \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c30\u0c42\u0c2a\u0c3e\u0c02\u0c24\u0c30 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c2e\u0c41 \u0c15\u0c28\u0c41\u0c15\u0c4d\u0c15\u0c4b\u0c02\u0c21\u0c3f. [Mar. ’14, May ’07]
\n\u0c38\u0c3e\u0c27\u0c28:
\n\u0c26\u0c24\u0c4d\u0c24 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c2e\u0c41 x cos \u03b1 + y sin \u03b1 = p
\n\u2235 \u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c15\u0c3e\u0c15\u0c4d\u0c37\u0c3e\u0c32\u0c28\u0c41 \u0c15\u0c4b\u0c23\u0c02 \u0c2d\u0c4d\u0c30\u0c2e\u0c23\u0c02 \u0c1a\u0c47\u0c38\u0c3e\u0c2f\u0c3f.
\nx = x’ cos \u03b1 – y’ sin \u03b1
\ny = x’ sin \u03b1 + y’ cos \u03b1
\n\u0c26\u0c24\u0c4d\u0c24 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c2e\u0c41 \u0c2a\u0c30\u0c3f\u0c35\u0c30\u0c4d\u0c24\u0c3f\u0c24 \u0c30\u0c42\u0c2a\u0c2e\u0c41
\n(x’ cos \u03b1 – y’ sin \u03b1) cos \u03b1 + (x’ sin a + y’ cos \u03b1) sin \u03b1 = p
\n\u21d2 x’ (cos2<\/sup> \u03b1 + sin2<\/sup> \u03b1) = p \u21d2 x’ = p
\n\u0c26\u0c24\u0c4d\u0c24 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c2e\u0c41 \u0c2a\u0c30\u0c3f\u0c35\u0c30\u0c4d\u0c24\u0c3f\u0c24 \u0c30\u0c42\u0c2a\u0c02 x = p<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 4.
\n\\(\\frac{\\pi}{6}\\) \u0c15\u0c4b\u0c23\u0c02\u0c24\u0c4b \u0c05\u0c15\u0c4d\u0c37\u0c3e\u0c32\u0c28\u0c41 \u0c2d\u0c4d\u0c30\u0c2e\u0c23\u0c02 \u0c1a\u0c47\u0c38\u0c3f\u0c28\u0c2a\u0c4d\u0c2a\u0c41\u0c21\u0c41, x2<\/sup> + 2\\(\\sqrt{3}\\)xy – y2<\/sup> = 2a2<\/sup> \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c2e\u0c41 \u0c15\u0c28\u0c41\u0c15\u0c4d\u0c15\u0c4b\u0c02\u0c21\u0c3f. [A.P Mar. ’15, ’12, ’07, ’04; May ’13, ’12]
\n\u0c38\u0c3e\u0c27\u0c28:
\n\u03b8 = \\(\\frac{\\pi}{6}\\), x = X cos \u03b1 + Y sin \u03b1
\nx = X cos \\(\\frac{\\pi}{6}\\) – Y sin \\(\\frac{\\pi}{6}\\)
\n= \\(X \\cdot \\frac{\\sqrt{3}}{2}-Y \\cdot \\frac{1}{2}=\\frac{\\sqrt{3} X-Y}{2}\\)
\ny = X sin \u03b1 + Y cos \u03b1 = X. sin \\(\\frac{\\pi}{6}\\) + Y cos \\(\\frac{\\pi}{6}\\)
\n= \\(X \\cdot \\frac{1}{2}+Y . \\frac{\\sqrt{3}}{2}=\\frac{X+\\sqrt{3} Y}{2}\\)
\n\u0c30\u0c42\u0c2a\u0c3e\u0c02\u0c24\u0c30 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c2e\u0c41
\n\"AP
\n\u21d2 \\(\\frac{3 x^2-2 \\sqrt{3} X Y+Y^2}{4}\\) + \\(\\frac{2 \\sqrt{3}\\left[\\sqrt{3} X^2-X Y+3 X Y-\\sqrt{3} Y^2\\right]}{4}\\) – \\(\\frac{X^2+3 Y^2+2 \\sqrt{3} X Y}{4}\\) = 2a2<\/sup>
\n\u21d2 3X2<\/sup> – 2\\(\\sqrt{3}\\) XY + Y2<\/sup> + 2\\(\\sqrt{3}\\) [\\(\\sqrt{3}\\)X2<\/sup> + 2XY – \\(\\sqrt{3}\\)Y2<\/sup>] – (x2<\/sup> + 3Y2<\/sup> + 2\\(\\sqrt{3}\\)XY) = 8a2<\/sup>
\n\u21d2 3x2<\/sup> – 2\\(\\sqrt{3}\\)XY + Y2<\/sup> + 6X2<\/sup> + 4\\(\\sqrt{3}\\)XY – 6Y2<\/sup> – X2<\/sup> – 3Y2<\/sup> – 2\\(\\sqrt{3}\\) XY = 8a2<\/sup>
\n\u21d2 8x2<\/sup> – 8y2<\/sup> = 8a2<\/sup> \u21d2 X2<\/sup> – Y2<\/sup> = a2<\/sup><\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 5.
\n\\(\\frac{\\pi}{4}\\) \u0c15\u0c4b\u0c23\u0c02\u0c24\u0c4b \u0c05\u0c15\u0c4d\u0c37\u0c3e\u0c32\u0c28\u0c41 \u0c2d\u0c4d\u0c30\u0c2e\u0c23\u0c02 \u0c1a\u0c47\u0c38\u0c3f\u0c28\u0c2a\u0c4d\u0c2a\u0c41\u0c21\u0c41, 3x2<\/sup> + 10xy + 3y2<\/sup> = 9 \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c30\u0c42\u0c2a\u0c3e\u0c02\u0c24\u0c30
\n\u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c2e\u0c41 \u0c15\u0c28\u0c41\u0c15\u0c4d\u0c15\u0c4b\u0c02\u0c21\u0c3f. [May ’11]
\n\u0c38\u0c3e\u0c27\u0c28:
\n\u0c26\u0c24\u0c4d\u0c24 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c2e\u0c41
\n3x2<\/sup> + 10xy + 3y2<\/sup> – 9 = 0 ……………… (1)
\n\u0c2d\u0c4d\u0c30\u0c2e\u0c23 \u0c15\u0c4b\u0c23\u0c2e\u0c41 \u03b8 = \\(\\frac{\\pi}{4}\\)
\n(X, Y) \u0c32\u0c41 (x, y) \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c28\u0c42\u0c24\u0c28 \u0c28\u0c3f\u0c30\u0c42\u0c2a\u0c15\u0c3e\u0c32\u0c41 \u0c05\u0c28\u0c41\u0c15\u0c4a\u0c28\u0c41\u0c2e\u0c41.
\nx = X cos \u03b8 – Y sin \u03b8
\n= X cos \\(\\frac{\\pi}{4}\\) – y sin \\(\\frac{\\pi}{4}\\) = \\(\\frac{X-Y}{\\sqrt{2}}\\)
\ny = X sin \u03b8 + Y cos \u03b8 = X sin \\(\\frac{\\pi}{4}\\) + Y cos \\(\\frac{\\pi}{4}\\) = 2 \\(\\frac{X+Y}{\\sqrt{2}}\\)
\n(1) \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c30\u0c42\u0c2a\u0c3e\u0c02\u0c24\u0c30 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c2e\u0c41
\n\\(3\\left(\\frac{X-Y}{\\sqrt{2}}\\right)^2+10\\left(\\frac{X-Y}{\\sqrt{2}}\\right)\\left(\\frac{X+Y}{\\sqrt{2}}\\right)+3\\left(\\frac{X+Y}{\\sqrt{2}}\\right)^2-9=0\\)
\n\u21d2 3 \\(\\frac{\\left(X^2-2 X Y+Y^2\\right)}{2}\\) + 10 \\(\\frac{\\left(X^2-Y^2\\right)}{2}\\) + 3 \\(\\frac{\\left(X^2+2 X Y+Y^2\\right)}{2}\\) – 9 = 0
\n\u21d2 3x2<\/sup> – 6XY + 3y2<\/sup> + 10X2<\/sup> – 10Y2<\/sup> + 3X2 <\/sup>+ 6XY + 3Y2<\/sup> – 18 = 0
\n16X2<\/sup> – 4Y2<\/sup> – 18 = 0
\n\u2234 8X2<\/sup> – 2Y2<\/sup> = 9
\n\u2234 8X2<\/sup> – 2Y2<\/sup> = 9<\/p>\n

\"AP<\/p>\n","protected":false},"excerpt":{"rendered":"

Practicing the\u00a0Intermediate 1st Year Maths 1B Textbook Solutions Chapter 2 \u0c05\u0c15\u0c4d\u0c37 \u0c2a\u0c30\u0c3f\u0c35\u0c30\u0c4d\u0c24\u0c28\u0c02 Exercise 2(a) will help students to clear their doubts quickly. AP Inter 1st Year Maths 1B Solutions Chapter 2 \u0c05\u0c15\u0c4d\u0c37 \u0c2a\u0c30\u0c3f\u0c35\u0c30\u0c4d\u0c24\u0c28\u0c02 Exercise 2(a) \u0c05\u0c2d\u0c4d\u0c2f\u0c3e\u0c38\u0c02 \u2013 2 (\u0c0e) I. \u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 1. \u0c05\u0c15\u0c4d\u0c37\u0c3e\u0c32 \u0c38\u0c2e\u0c3e\u0c02\u0c24\u0c30\u0c2a\u0c30\u0c3f\u0c35\u0c30\u0c4d\u0c24\u0c28 \u0c26\u0c4d\u0c35\u0c3e\u0c30\u0c3e \u0c2e\u0c42\u0c32\u0c2c\u0c3f\u0c02\u0c26\u0c41\u0c35\u0c41\u0c28\u0c41 (4, \u22125) \u0c15\u0c41 \u0c2e\u0c3e\u0c30\u0c3f\u0c38\u0c4d\u0c24\u0c47 \u0c15\u0c4a\u0c24\u0c4d\u0c24 \u0c05\u0c15\u0c4d\u0c37\u0c3e\u0c32 \u0c26\u0c43\u0c37\u0c4d\u0c1f\u0c4d\u0c2f\u0c3e \u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c2c\u0c3f\u0c02\u0c26\u0c41\u0c35\u0c41\u0c32 … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[14],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/29255"}],"collection":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/comments?post=29255"}],"version-history":[{"count":4,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/29255\/revisions"}],"predecessor-version":[{"id":30124,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/29255\/revisions\/30124"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=29255"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=29255"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=29255"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}