slopes of parallel lines are equal. Consider two parallel lines AB - PowerPoint PPT Presentation

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D AY 28 A PPLICATION OF SLOPE CRITERIA OF SPECIAL LINES I NTRODUCTION We have learned that two parallel lines have equal slopes. We have also learned that the product of slopes of two perpendicular line is -1. In this lesson, we will

D AY 28 – A PPLICATION OF SLOPE CRITERIA OF SPECIAL LINES
I NTRODUCTION We have learned that two parallel lines have equal slopes. We have also learned that the product of slopes of two perpendicular line is -1. In this lesson, we will discuss how we can apply this knowledge to find the equation of the line which is perpendicular or parallel to a given line that passes through a given point.
V OCABULARY  Slope of a line A number that measures steepness of a line.
When given two parallel lines and two points on one line which can help us to find the slope of the line, we can find the slope of the other line since slopes of parallel lines are equal. Consider two parallel lines AB and CD below. The coordinates of points A, B, and C are given, but the coordinates of D are not given. B( 𝑦 2, 𝑧 2 ) D A( 𝑦 1, 𝑧 1 ) C( 𝑏, 𝑐 )
To find the equation of line CD we first find the slope of line AB. 𝑧 2 −𝑧 1 Slope of CD = Slope of AB = 𝑦 2 −𝑦 1 After getting the slope of line CD, we then use the coordinates of point C to find equation of the line CD as follows; 𝑧−𝑐 𝑧 2 −𝑧 1 𝑧−𝑐 𝑦−𝑏 = 𝑦 2 −𝑦 1 (equate 𝑦−𝑏 with the slope of CD)
 Example Lines ST and MN are parallel. Line MN passes through points 𝑁 4,5 and 𝑂 2,2 . If line ST passes through point 𝑇 3,1 f ind the equation of line ST. Solution 5−2 3 Slope of MN = 4−2 = 2 3 Slope of ST = Slope of MN = 2 𝑧−1 3 2 (to find the equation of line ST) 𝑦−3 = 2𝑧 − 2 = 3𝑦 − 9 3 7 2 x − 𝑧 = 2
Consider two perpendicular lines JK and KL below which intersect at point K. We are given the coordinates of points J and K. J 𝑦 2 , 𝑧 2 K 𝑦 1 , 𝑧 1 L We can find the slope of JK since we have the coordinates of J and K. 𝑧 2 −𝑧 1 Slope of JK = 𝑦 2 −𝑦 1 Slope of JK × Slope of KL = −1 𝑧 2 −𝑧 1 𝑦 2 −𝑦 1 × Slope of KL = −1 𝑦 2 −𝑦 1 Slope of KL = − 𝑧 2 −𝑧 1
Example Two perpendicular lines HI and IJ intersect at point I 4,8 . Line HI passes through point H 2,4 . Find the equation of line IJ. Solution 8−4 Slope of HI = 4−2 = 2 Slope of HI × Slope of IJ = -1 2 × Slope of I J = -1 1 Slope of I J = − 2 𝑧−8 1 𝑦−4 = − 2 2𝑧 − 16 = −𝑦 + 4 1 𝑧 = − 2 𝑦 + 10
HOMEWORK Two perpendicular lines AB and AC intersect at point A 4,6 . Line AB passes through point B 5,4 . Find the equation of line AC.
A NSWERS TO HOMEWORK 1 𝑧 = 2 𝑦 + 4
THE END

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