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n t U a sk Sa 2001 Doug MacLean x y 5 4 3 2 1 0 -1 -3-2-101234 5 = 25 √ = 2 = ) 1 9 . y ) is + 4 1 1 − ) , 2 y x 2 3 16 ( √ − − ,y (y 2 2 2 = + x y (x 2 2 to be 3 ) = = 1 ) + 2 2 x x y 2 P 1 − ∆ 4 ∆ ,y 2 2 is = ) 2 and (x = (x 2 ) ) 1 2 = 1 ) ,y y run rise 2 2 1 ,y P − |= 1 = − 2 (x 2 P 4 (x 1 = ( m (y and P 2 | = + ) + P 1 1 2 2 , P ) )) 1 1 1 and − x ( − ) ( − 1 2 Slope − ,y 1 3 (x 3 4 to be ( (x ) 1 = |= = ,y 1 2 1 ) P P 1 1 Coordinate Geometry & Lines 1 (x P − | ( − = 4 1 − P 3 = m In our diagram, we take Distance Formula It is essential that the student be able to automatically apply the very basicThe formulas distance of between elementary the Analytic points Geometry: Then The slope of the line passing through the points so in our example
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n t U a sk Sa 2001 Doug MacLean x x x y y 4 3 2 1 0 -1 4 3 2 1 0 -1 y 4 3 2 1 0 -1 -3-2-101234 -2-1012345 -3-2-101234 = 1 1. y = − 3 y is y is + ) m 4 x 2 2 + ) is ,y 1 2 is x x m ) 1 2 (x 3 − , = is 0 =− 2 (x ( with slope 1 P 1 y ,b) y x 0 and ,b) ( − − 2 and 0 ) 2 2 with slope ( ) 0 x y 1 ) , and 1 4 = ( ,y ) 1 ,y 1 0 1 y (x (x (a, − the equation of the line is = = y 1 1 3 1 P P 2, the equation is =− = -axis at the point m y b and 1 2 Equations of Lines =− and the slope m ) 2 , 1 ( = 1 P ) 1 − . For example, if (x b 1 3 ) 1. For example, the equation of the line through + 1 = x =− − b y mx 2 + = − Point-Slope Form Point-Point Form Slope-Intercept Form Intercept-Intercept Form a x y m(x This just comes from putting the two previous formulas together. y Thus given the point These come in many useful forms: The equation of the line passing through the point The equation of the line passing through the points The equation of the line passing through the The equation of the line passing through the intercepts
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n t U a sk Sa 2001 Doug MacLean 2. + x 2 1 =− y . 0 x = which is perpendicular to the line C ) 1 + , 3 2 2, so the equation of the perpendicular line is, using the Point-Slope ( By = . + 1 2 2 1 − m Ax − = y 1 4 3 2 1 0 -1 m -3-2-101234 parallel if are 2 Parallel & Perpendicular Lines m 1. and =− ) 1 2 2 m − m 1 (x m 2 Find the equation of the line through the point The slope of the perpendicular line is = 1 − y General Form Example: Solution: perpendicular if Form: Every line has infinitely many equations of the form For any fixed line, they are non-zero multiples of each other. Two lines with slopes
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n t U a sk = Sa 2001 Doug MacLean B 1, = A 4, = 0 y 3, = 0 x 2. + x 2 1 =− y 5 9 √ 4 = | 2 0, so we apply the Distance Formula with 4 − to the line with equation + = 8 ) 1 2 4 + √ , with equation − 3 3 | ( y 2 = | + ) 2 x − ( to a line 2 ) + ) 0 2 ) ( 2or 4 ,y x 0 + + )( 2 2 x (x ( ) 1 = + ( Distance from a Point to a Line ) 0 =− 3 P y )( 1 ( | = | | C y C Find the distance from the point We must rewrite the equation of the line in General form: + 2 + 2 0is 4 3 2 1 0 -1 0 B 0 B = + + By 2. becomes 2 By 2 C 2 2: + A + + A + 0 √ 0 √ x -3-2-101234 =− 2 1 Ax By Ax | | C + = =− = Example: Solution: d Ax d The distance from the point y and