0000004364 00000 n %PDF-1.3 %���� 0000005278 00000 n 0000002291 00000 n 0000001914 00000 n 0000004564 00000 n 0000003253 00000 n 0000003334 00000 n 0000011229 00000 n 0000012954 00000 n 0000014389 00000 n Answers & Solutions: 1) Answer (C) Let line $l$ perpendicularly bisects line joining A(2,-5) and B(0,7) at C, thus C is the mid point of AB. 1 Which of the following is the correct statement of Pythagoras’ theorem for the triangle shown? 0000014994 00000 n Thsese solutions are very easy to understand. 0000015133 00000 n 33 0 obj << /Linearized 1 /O 35 /H [ 1358 352 ] /L 109748 /E 84443 /N 3 /T 108970 >> endobj xref 33 43 0000000016 00000 n 0000029667 00000 n 0000001358 00000 n 0000012933 00000 n 0000011400 00000 n 0000008448 00000 n 0000003185 00000 n 0000061192 00000 n 0000013478 00000 n 0000001689 00000 n Observe the example problems and solutions. 0000028512 00000 n Uses Pythagoras’ theorem to find 40 9 EP. Exercise 7.1, exercise 7.2, exercise 7.3, and exercise 7.4 solutions are given. 0000012473 00000 n 0000007827 00000 n 0000015388 00000 n 0000002096 00000 n 0000013499 00000 n 0000010467 00000 n 0000001693 00000 n Coordinate Geometry is the branch of mathematics that helps us to exactly locate a given point with the help of an ordered pair of numbers. 0000013958 00000 n (see a mnemonic for this formula) For example: The equation of the line in the above diagram is: y = ½ x - 1 How to Find the Slope Given 2 Points? 0000002309 00000 n trailer << /Size 71 /Info 16 0 R /Root 29 0 R /Prev 102210 /ID[] >> startxref 0 %%EOF 29 0 obj << /Type /Catalog /Pages 17 0 R /OpenAction [ 30 0 R /XYZ null null null ] /PageMode /UseNone /JT 27 0 R /PageLabels 15 0 R >> endobj 69 0 obj << /S 107 /T 216 /L 260 /Filter /FlateDecode /Length 70 0 R >> stream 0000050249 00000 n 0000001710 00000 n H�d�mHSQ���zg��޼ˈ�M Between points A and B: AB 2 = (Bx – Ax) 2 + (By – Ay) 2 The Midpoint of a Line Joining Two Points 0000007218 00000 n C1 COORDINATE GEOMETRY Answers - Worksheet D page 2 Solomon Press 7 a at A, y = 0 ∴ x = 20 8 a grad q = grad p = 3 4 − at B, x = 0 ∴ y = 10 ∴ y = 3 4 − x + 7 ∴ A (20, 0), B (0, 10) b grad r = 4 3 b l ⇒ y = 10 − 1 2 x ∴ y = 4 3 (x − 1) ∴ grad of l = 1 2 − 3y = 4x − 4 First study the text book very well. => Coordinates of C = $(\frac{2 + 0}{2} , \frac{-5 + 7}{2})$ = $(\frac{2}{2} , \frac{2}{2}) = (1,1)$ Now, slope of AB = $\frac{y_2 – y_1}{x_2 – x_1} = \frac{(7 + 5)}{(0 – 2)}$ = $\frac{12}{-2} = -6$ 0000002078 00000 n 0000008427 00000 n trailer << /Size 76 /Info 21 0 R /Root 34 0 R /Prev 108960 /ID[] >> startxref 0 %%EOF 34 0 obj << /Type /Catalog /Pages 22 0 R /OpenAction [ 35 0 R /XYZ null null null ] /PageMode /UseNone /JT 32 0 R /PageLabels 20 0 R >> endobj 74 0 obj << /S 142 /T 251 /L 296 /Filter /FlateDecode /Length 75 0 R >> stream 0000007833 00000 n 0000007613 00000 n 0000012452 00000 n 0000005839 00000 n 0000009162 00000 n 0000001207 00000 n 0000010235 00000 n 0000005962 00000 n 0000003036 00000 n 0000010257 00000 n 0000007432 00000 n Coordinate geometry is also said to be the study of graphs. 0000013937 00000 n 0000014410 00000 n The number plane is the basis of coordinate geometry, an important branch of mathematics. 0000005008 00000 n 0000011421 00000 n 0000009466 00000 n 0000002746 00000 n C1 COORDINATE GEOMETRY Answers - Worksheet A page 3 Solomon Press 16 2AB = 82 + 102 = 164 17 a PQ = 6222+ = 40 = 210 AB = 164 = 241 PR = 11722+ = 290 radius = 1 2 AB = 41 22QR = 515+ = 250 = 510 area = π 2× ( 41) = 41π b PQ2 + QR2 = 40 + 250 = 290 = PR2 ∴ by converse of Pythagoras’ ∠PQR is a right-angle c area = 1 2 × PQ × QR = 50 0000060983 00000 n 28 0 obj << /Linearized 1 /O 30 /H [ 1358 335 ] /L 102898 /E 85931 /N 2 /T 102220 >> endobj xref 28 43 0000000016 00000 n Complete a right angle triangle and use Pythagoras' theorem to work out the length of the line. 0000002968 00000 n 0000002814 00000 n ¨¸ ©¹ M1 2.2a 11 33 yx A1 1.1b (4) 5e PA 40 B1 3.1a 5th Solve coordinate geometry problems involving circles in context. 0000012098 00000 n 0000008218 00000 n 0000007002 00000 n Example: Find the slope of the two points (-6,3) and (4,-3) 0000005490 00000 n 0000009183 00000 n 0000001207 00000 n In figure ABC is a triangle coordinates of whose vertex A are (0, -1). Graphs are visual representations of our data. H�bf�cc��[email protected] AV�(�� �:t��_D�@�9E�O�f,�c���h�)� �[email protected]�,�i 0000076174 00000 n 0000010446 00000 n The Distance Between two Points. In this chapter, we will look at some of the basic ideas of coordinate geometry and how they can be used to solve problems. 0000012077 00000 n 0000007216 00000 n 0000008047 00000 n Draw a line between the two points. �lMj��#A��%��ns���[Z����ɚA��r4+��A�!R��W�KD_�P�s�g���˗����������9ef �1g����ZV[;[���c""�1)� Y�I H�bf�cc�,[email protected] AV�(��)�@����f~�S�+�xJ. 0000001932 00000 n 0000009445 00000 n This section looks at Coordinate Geometry. r���ޟ��9��"�g8��9Y�;Nͅ���ʷh�s�W�Z������Z\��vQjq��EItz%�&Iހ�ɝn�. 0000004715 00000 n D and E respectively are the mid-points of the sides AB and AC and their coordinates are (1, 0) and (0,1) respectively. 0000050458 00000 n 0000005635 00000 n 0000005984 00000 n 0000011250 00000 n 0000001358 00000 n 0000003402 00000 n 0000004211 00000 n If F is the mid-point of BC, find the areas of ∆ABC and ∆DEF. B1 2.2a Area of EPA = 1 40 40 29 uu SSC solutions for maths Coordinate Geometry Mathematics solutions for Coordinate Geometry class 10 SSC. 0000005423 00000 n A a2 = b2 + c2 B b2 = a2 + c2 C c2 = a2 + b2 0000005817 00000 n 0000015527 00000 n 0000074909 00000 n 0000004857 00000 n Coordinate geometry is the combination of geometry and algebra to solve the problems. @Z���"����l��\$�#�f�8�4pPs`p�f8��Y�kCqs#�� �@k650�W1ޜ�j�7'hc�\ƜL����X�4����C��������L][email protected]"|�]MPyo� %)q endstream endobj 70 0 obj 215 endobj 30 0 obj << /Type /Page /Parent 17 0 R /Resources 31 0 R /Contents [ 44 0 R 50 0 R 52 0 R 54 0 R 56 0 R 58 0 R 60 0 R 62 0 R ] /Thumb 7 0 R /MediaBox [ 0 0 595 842 ] /CropBox [ 0 0 595 842 ] /Rotate 0 >> endobj 31 0 obj << /ProcSet [ /PDF /Text ] /Font << /F2 35 0 R /TT2 33 0 R /TT4 40 0 R /TT6 42 0 R /TT8 46 0 R /TT10 48 0 R >> /ExtGState << /GS1 63 0 R >> >> endobj 32 0 obj << /Type /FontDescriptor /Ascent 891 /CapHeight 0 /Descent -216 /Flags 34 /FontBBox [ -568 -307 2028 1007 ] /FontName /AHMALG+TimesNewRoman /ItalicAngle 0 /StemV 0 /FontFile2 65 0 R >> endobj 33 0 obj << /Type /Font /Subtype /TrueType /FirstChar 32 /LastChar 121 /Widths [ 250 0 0 0 0 0 0 0 333 333 0 564 250 333 0 0 500 500 500 500 500 500 500 500 500 500 278 0 0 564 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 333 0 333 0 0 0 444 500 444 500 444 333 500 500 278 0 500 278 778 500 500 500 500 333 389 278 500 0 722 500 500 ] /Encoding /WinAnsiEncoding /BaseFont /AHMALG+TimesNewRoman /FontDescriptor 32 0 R >> endobj 34 0 obj << /Type /FontDescriptor /Ascent 891 /CapHeight 0 /Descent -216 /Flags 98 /FontBBox [ -498 -307 1120 1023 ] /FontName /AHMAPD+TimesNewRoman,Italic /ItalicAngle -15 /StemV 0 /FontFile2 67 0 R >> endobj 35 0 obj << /Type /Font /Subtype /Type1 /FirstChar 1 /LastChar 9 /Widths [ 790 549 549 863 549 987 612 768 400 ] /Encoding 36 0 R /BaseFont /AHMANA+Symbol /FontDescriptor 39 0 R /ToUnicode 38 0 R >> endobj 36 0 obj << /Type /Encoding /Differences [ 1 /copyrightserif /minus /plus /therefore /multiply /arrowdblright /Delta /angle /degree ] >> endobj 37 0 obj << /Filter /FlateDecode /Length 786 /Subtype /Type1C >> stream In coordinate geometry, the equation of a line can be written in the form, y = mx + b, where m is the slope and b is the y-intercept. %PDF-1.3 %���� 0000008197 00000 n 0000001672 00000 n