If F is the mid-point of BC, find the areas of ∆ABC and ∆DEF. 0000001710 00000 n 0000010257 00000 n 0000012077 00000 n 0000004564 00000 n 0000013499 00000 n 0000012452 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 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 H�b```f``�c`c`��dd@ AV�(�� �:t��_D�@�9E�O�f,�c`���h�)� �g@�,�i 0000004715 00000 n Coordinate geometry is also said to be the study of graphs. H�d�mHSQ���zg��޼ˈ�M %PDF-1.3 %���� (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? Draw a line between the two points. Between points A and B: AB 2 = (Bx – Ax) 2 + (By – Ay) 2 The Midpoint of a Line Joining Two Points Thsese solutions are very easy to understand. In figure ABC is a triangle coordinates of whose vertex A are (0, -1). 0000001358 00000 n 0000007218 00000 n H�b```f``�c`c`�,ed@ AV�(��)�@����f~�S�+�xJ. 0000007216 00000 n 0000007827 00000 n 0000002968 00000 n B1 2.2a Area of EPA = 1 40 40 29 uu 0000076174 00000 n => Coordinates of C = $(\frac{2 + 0}{2} , \frac{-5 + 7}{2})$ = $(\frac{2}{2} , \frac{2}{2}) = (1,1)$ Now, slope of AB = $\frac{y_2 – y_1}{x_2 – x_1} = \frac{(7 + 5)}{(0 – 2)}$ = $\frac{12}{-2} = -6$ Observe the example problems and solutions. 0000008427 00000 n 0000011421 00000 n 0000005423 00000 n 0000009466 00000 n 0000014410 00000 n First study the text book very well. 0000005490 00000 n Exercise 7.1, exercise 7.2, exercise 7.3, and exercise 7.4 solutions are given. 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 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. 0000011229 00000 n 0000012098 00000 n 0000013958 00000 n The number plane is the basis of coordinate geometry, an important branch of mathematics. 0000005839 00000 n 0000003036 00000 n 0000001358 00000 n %PDF-1.3 %���� 0000014389 00000 n 0000008218 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 0000005278 00000 n Graphs are visual representations of our data. 0000012933 00000 n 1 Which of the following is the correct statement of Pythagoras’ theorem for the triangle shown? SSC solutions for maths Coordinate Geometry Mathematics solutions for Coordinate Geometry class 10 SSC. �lMj��#A��`%��ns���[Z����ɚA��r4+��A�!R��W�KD_�P�s�g���˗����������9ef �1g����ZV[;[���c""�1)� Y�I 0000005962 00000 n 0000002814 00000 n 0000009183 00000 n 0000001914 00000 n 0000005817 00000 n 0000007002 00000 n 0000004364 00000 n 0000003402 00000 n 0000004857 00000 n 0000061192 00000 n 0000013478 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 0000004211 00000 n 0000008047 00000 n 0000002746 00000 n This section looks at Coordinate Geometry. Coordinate geometry is the combination of geometry and algebra to solve the problems. 0000011400 00000 n Example: Find the slope of the two points (-6,3) and (4,-3) 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. 0000010446 00000 n 0000012473 00000 n 0000029667 00000 n 0000015527 00000 n 0000060983 00000 n A a2 = b2 + c2 B b2 = a2 + c2 C c2 = a2 + b2 0000002096 00000 n 0000002309 00000 n 0000003185 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. 0000005008 00000 n 0000012954 00000 n In this chapter, we will look at some of the basic ideas of coordinate geometry and how they can be used to solve problems. 0000009162 00000 n 0000015388 00000 n 0000005984 00000 n 0000014994 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 0000009445 00000 n 0000003334 00000 n 0000011250 00000 n 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. Complete a right angle triangle and use Pythagoras' theorem to work out the length of the line. 0000008448 00000 n @Z���"����l��$�#�f�8�4pPs`p�f8��Y�kCqs#�� �@k650�W1ޜ�j�7'hc�\ƜL����X�4����C��������L]+@"|�]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 0000074909 00000 n 0000050458 00000 n 0000001693 00000 n 0000008197 00000 n The Distance Between two Points. 0000007613 00000 n 0000001672 00000 n 0000028512 00000 n 0000002078 00000 n 0000001207 00000 n r���ޟ��9��"�g8��9Y�;Nͅ���ʷh�s�W�Z������Z\��vQjq��EItz%�&Iހ�ɝn�. 0000007833 00000 n 0000010235 00000 n 0000005635 00000 n Uses Pythagoras’ theorem to find 40 9 EP. D and E respectively are the mid-points of the sides AB and AC and their coordinates are (1, 0) and (0,1) respectively. 0000050249 00000 n 0000001932 00000 n 0000002291 00000 n 0000001689 00000 n 0000001207 00000 n
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