THE EQUATION OF A LINE By: Mr. F. A. Ogrimen Jr..

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Presentation transcript:

THE EQUATION OF A LINE By: Mr. F. A. Ogrimen Jr.

Equation of a Line: A. Using the Slope-Intercept Form, y = mx + b where m is the slope and b is the y-intercept If the given data are slope and y-intercept, we can use the formula, y = mx + b. Examples: Find the equation of a line with a slope = -3 and y-intercept = 5. Solution: use the pattern, y = mx + b. Given: m=-3 and b=5; substitute to the formula to get, m b y = x To write the standard form of the equation, simply transpose the mx value to the left and change the sign. y = 53x +

Equation of a Line: B. Using the Point - Slope Form, y – y 1 = m(x – x 1 ) where (x 1, y 1 ) is the point and m is the slope If the given data are the point and the slope, we can use the formula, y – y 1 = m(x – x 1 ). Examples 1: Find the equation of the line with a slope = -3 and passes through ( 4, 2). Solution: use the pattern, y – y 1 = m(x – x 1 ). Given: m=-3 and x 1 = 4, y 1 = 2; substitute to the formula to get, m y1 y1 Simplify: y – 2 = -3x + 12 y – = (x – ) x1 x x + y = x + y = 14

Equation of a Line: B. Using the Point - Slope Form, y – y 1 = m(x – x 1 ) where (x 1, y 1 ) is the point and m is the slope Examples 1: Find the equation of the line with a slope = and passes through ( 4, –5 ). m - y 1 Simplify: y = (x – ) x1 x1 Solution: use the pattern, y – y 1 = m(x – x 1 ). Given: and x 1 = 4, y 1 = -5; substitute to the formula to get, y + 15 = -2x + 8 2x + 3y = 8 – 15 2x + 3y = –7

Equation of a Line: How about if the given are Two Points, P 1 (x 1, y 1 ) ; P 2 (x 2, y 2 ) ? Examples 1: Find the equation of a line passing through the points, P 1 (3, 8) ; P 2 (1, 2). C.You can still use the Point - Slope Form, y – y 1 = m(x – x 1 ) but 1 st. Solve for slope, 2 nd. Apply the point-slope formula, y – y 1 = m(x – x 1 ) Solution: 1 st. 2 nd. Use the formula y – 8 = 3(x – 3) y – 8 = 3x – 9 – 3x + y = –9 + 8 – 3x + y = –1 or 3x – y = 1

Equation of a Line: Observe these Two Points, (a, 0) ; (0, b) Examples 1: Find the equation of a line passing through the points, (3, 0) and (0, –4 ). The two points are the x and y - intercepts D.Very easy to find the equation of a line. Simply apply the formula, Solution: To eliminate the denominator, simply multiply both sides by the LCD which is 12: 4x – 3y = 12:

Equation of a Line: Observe these Two Points, (a, 0) ; (0, b) Examples 2: Find the equation of a line passing through the x and y -intercepts, The two points are the x and y - intercepts D.Very easy to find the equation of a line. Simply apply the formula, Solution: Since the denominator is fraction, multiply its reciprocal to the numerator: then multiply both side by the LCD, 35 21x + 10y = 35:

E. Equation of a Horizontal Line: y = b If from the given points the value of y is repeated. (2, b) ; (0, b) ; (-3, b) Answer: y = 6 Example 1) Find the equation of a line passing through the points: ( 5, 6 ) ; ( -3, 6) F. Equation of a Vertical Line: x = a If from the given points the value of x is repeated. (a, 5) ; (a, -3) ; (a, 1) Example 1) Find the equation of a line passing through the points: ( -4, 6 ) ; ( -4, 6) Answer: x = -4