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Co-Ordinate Geometry Maths Studies

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**Distance Learning Objectives**

To find the distance of a line segment using two co-ordinates. To find coordinates when given 1 coordinate and the distance value. Determine the shape of a shape e.g. triangle using the distance formula.

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Distance Example 1: Use the Distance Formula to find the distance between the points with coordinates (−3, 4) and (5, 2).

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**Distance √(-1- -2)²+(q – 4)² = √10**

Example Find q given that P(-2,4) and Q(-1,q) are √10 units apart. √(-1- -2)²+(q – 4)² = √10 √(-1)² + (q – 4)² = √10 Square both sides to remove √ (-1)² + (q – 4)² = 10 1 + (q-4)² = 10 (q-4)² = 10 – 1 (q – 4)² = 9 (q – 4)² = √9 To remove the √ find the √ of the other side q – 4 = 3 q = 7

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**Midpoint Learning Objective**

To find the midpoint of a line segment using two co-ordinates. To find coordinates when given 1 coordinate and the midpoint.

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Midpoint Find the midpoint R of the line segment with coordinates (-9,-1) and (-3,7)

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Midpoint

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Gradient The gradient of a line is a measurement of the steepness and direction of a nonvertical line.

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**Gradient Learning Objectives**

To find the gradient of a line given two coordinates. To find coordinates when given 1 coordinate and the gradient. Understand the relationship of parallel and perpendicular lines. Find the gradient of parallel and perpendicular lines

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**Gradient Formula Example M = Y2 – Y1 X2 – X1 A(1,5) B(2,8)**

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Drawing a Line Draw a line that passes through the point A(1,4) with gradient -1 (see pg. 91 eg 2)

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**Gradient Perpendicular Lines**

Perpendicular lines have opposite gradients. Eg Gradient of AB = 2/3 Gradient of CD = -3/2

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**Gradient Parallel Lines Parallel lines have the SAME gradient**

e.g. Gradient of MN = 3/9 Gradient of OP = 1/3

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Lines To prove that two lines are perpendicular, simply multiply their gradients. If the answer is -1, then the lines are perpendicular. Eg. The gradient of line AB is 2/3 The gradient of line CD is -3/2 2 x - 3

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Equation of a line An equation of a line can come in different forms depending on their gradient, if they are horizontal or vertical and if they intercept the y-axis.

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**Equation of a line Example 1 – Gradient Intercept Form y = mx + c**

Where m = gradient and c is the y-intercept y = mx + c (0,c) intercept See Example 5 P. 95

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**Equation of a Line Example 2 – General Form ax + by + c = 0**

To find the equation of a line in this form you need a gradient and one point. However, to find a gradient you need two points.

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Equation of a line E.g. Find the equation of a line that joins the points A(-3,5) B(1,2) The gradient is m = 2 – 5 = - 3 1-(-3) Using this formula y – y1=m(x – x1) y - -3 = -3/4(x – 5) 4(y + 3) = -3(x – 5) 4y + 12 = -3x + 15 3x + 4y – 3 = 0

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**Equation of a Line y - -3 = -3/4(x – 5) 4(y + 3) = -3(x – 5)**

Ans = 3x + 4y – 3 = 0

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1.4: equations of lines M(G&M)–10–9 Solves problems on and off the coordinate plane involving distance, midpoint, perpendicular and parallel lines, or.

1.4: equations of lines M(G&M)–10–9 Solves problems on and off the coordinate plane involving distance, midpoint, perpendicular and parallel lines, or.

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