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NOTE: A great way to study is to turn to the examples given in the text and try to solve them without looking at how it is done in the text. Once you have finished working through the problem by yourself, look through the solution outlined in your text book!

Chapter 2 Review – Analytic Geometry

2.1 Midpoint of a Line Segment (median, parallel & perpendicular lines, right bisectors)

|CONCEPTS |EXAMPLE PROBLEMS |

|To find the midpoint, M, of a line segment joining [pic]and [pic], use the |For the line segment ST, S has coordinates (6, 2) and T is on the y-axis. The|

|midpoint formula, [pic]. |midpoint, M, of ST is on the x-axis. Find the coordinates of T and M. |

| | |

|Parallel Lines – The slopes of 2 parallel lines are the same. The y-intercepts | |

|may be different or the same. |Determine the equation of the right bisector of a line with end points at (-2,|

|Perpendicular Lines – The slopes of 2 perpendicular lines are negative |5) and (-4, -3). |

|reciprocals. This means that when they are multiplied together, the answer is | |

|-1. | |

| |Determine the equation of the median from vertex A to midpoint BC of [pic] |

|Bisector – A bisector is a line which intersects another line at its midpoint. |with vertices at A(-1, 8), |

| |B(-4, 2), & C(0, -2). |

|Right Bisector – A right bisector is a line which intersects another line at its| |

|midpoint at a 90° angle. | |

| |[pic]has vertices A(3, 4), B(-5, 2), and C(1, -4). Determine an equation for:|

| |CD, the median from C to AB |

| |GH, the right bisector of AC |

|To find the equation of a right bisector: |2. Find the slope of AC: |

|[pic]has vertices A(3, 4), B(-5, 2), and C(1, -4). |A(3, 4) & C(1, -4) |

|Determine the equation of GH, the right bisector of AC |[pic] |

| | |

| |3. Find the equation of right bisector: |

| |G(2,0) mGH= -¼ (neg. rec. of mAC) |

| |[pic] OR [pic] so, y = -¼x + ½ |

| | |

|1. Find the midpoint, G, of AC: | |

|[pic] | |

| | |

| | |

|To find the equation of a median of a triangle: | |

|[pic]has vertices A(3, 4), B(-5, 2), and C(1, -4). Determine an equation for | |

|CD, the median from C to AB | |

| | |

| |3. Find equation of median using point-slope form or slope - y-intercept |

| |form: |

| |CDm = [pic] C(1, -4) |

| |[pic] OR [pic] |

|1. Find midpoint, D, of side AB: | |

|[pic] | |

| | |

|2. Find slope of median using points A & D: | |

|[pic] | |

|See examples on pages 60(find midpoint), 62 (eq. for median of triangle), 63(Eq. for right bisector). |

|Key concepts on page 65. |

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2.2 Length of a Line Segment

|CONCEPTS |EXAMPLE PROBLEMS |

|To find the length of a line segment joining [pic]and [pic], use the formula |Determine the perimeter of the quadrilateral with vertices A(3,2), B(-2, 4), |

|[pic]. |C(-4, -2), and D(4, -1) |

| |(a) as an exact solution |

| |(b) as an approximate solution to the nearest tenth. |

|Key Concepts page 76. |See examples on pages 74-76. |

2.3 Apply Slope, Midpoint, and Length Formulas (Shortest distance from a point to a line)

|CONCEPTS |EXAMPLE PROBLEMS |

|Determine equations for lines. |The equation of a circle with centre [pic] is [pic]. The points M(1,4) and |

|Determine when an angle is a right angle. |N(4, -1) are endpoints of chord MN. Find the distance from the centre of the |

|Classify triangles by side length (equilateral, isosceles, scalene). |circle to the chord, to the nearest tenth. |

|Determine type of quadrilateral by slopes & length of lines (parallelogram, | |

|rectangle). |[pic]has vertices A(3, 4), B(-5, 2), and C(1, -4). Determine an equation for |

| |AE, the altitude from A to BC. |

|To determine the equation of an altitude: | |

|Determine the equation of AE, the altitude from A to BC | |

|1. Find the slope of BC: | |

|B(-5, 2) & C(1, -4) |2. Find equation of altitude using point-slope form: |

|[pic] |A(3, 4) and negative reciprocal of mBC |

| |[pic] |

| |OR |

| |Use y=mx+b: |

| |[pic] so, y = x + 1 |

|To determine the shortest distance from a point to a line: | |

|To determine the distance from a point, P, to a line, AB, when given the | |

|coordinates of P, and the equation of AB: |You now have two equations for lines that intersect. Solve this linear system|

|Determine the slope of AB in order to find the slope of a perpendicular line. |using substitution or elimination. |

|Negative reciprocal. |equation of line AB which was given |

|Determine the equation of the perpendicular line to AB that passes through your |equation just found |

|point P using the point-slope form of a line. |FINALLY – find the length of the line segment between point P, and the point |

|[pic] |of intersection which you just found. |

| |[pic] |

|Key Concepts page 88. |See examples on pages 82-87. |

2.4 Equation for a Circle

|CONCEPTS |EXAMPLE PROBLEMS |

|An equation of the circle with centre [pic] and radius [pic]is [pic]. |Determine the radius of a circle given its centre (0, 0) if a point on its |

| |circumference is (2, 7). Round the radius to the nearest tenth, if necessary.|

|(More circles in section 3.5) |Write an equation for the circle. |

|Key Concepts page 96. |See examples on pages 94, 95. |

See Also

|Review on pages 100 to 103. |

|Chapter Test on pages 104 and 105. |

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