Graphing circles is a fairly simple process once we

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Unformatted text preview: iprocal of m1 . Example 4 Determine if the line that passes through the points ( -2, -10 ) and ( 6, -1) is parallel, perpendicular or neither to the line given by 7 y - 9 x = 15 . Solution Okay, in order to do answer this we’ll need the slopes of the two lines. Since we have two points for the first line we can use the formula for the slope, m1 = -1 - ( -10 ) 9 = 6 - ( -2 ) 8 We don’t actually need the equation of this line and so we won’t bother with it. Now, to get the slope of the second line all we need to do is put it into slope-intercept form. 7 y = 9 x + 15 9 15 y = x+ 7 7 Þ m2 = 9 7 Okay, since the two slopes aren’t the same (they’re close, but still not the same) the two lines are not parallel. Also, æ 9 öæ 9 ö 81 ¹ -1 ç ÷ç ÷ = è 8 øè 7 ø 56 so the two lines aren’t perpendicular either. Therefore, the two lines are neither parallel nor perpendicular. Example 5 Determine the equation of the line that passes through the point ( 8, 2 ) and is, (a) parallel to the line given by 10 y + 3 x = -2 [Solution] (b) per...
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