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## The Gradient Formula

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**The Gradient Formula**• Gradient of AB • = mAB • = height/horizontal • = BC/AC • = (y2-y1) (x2-x1) B(x2,y2) y2-y1 A(x1,y1) x2-x1 C(x2,y1)**The Gradient Formula ctd.**• Summary • If A is (x1,y1) and B is (x2,y2) • then the gradient of AB is given by • mAB = (y2-y1) (x2-x1)**Example 4**P is (5, 9) and Q is (7,17). • mPQ = (y2-y1) (x2-x1) • = (17 - 9) • ( 7 - 5) • = 8/2 • = 4 NB: line looks like Q P**Example 5**E is (15, 9) and F is (7,17). • mEF = (y2-y1) (x2-x1) • = (17 - 9) • ( 7 - 15) • = 8/-8 • = -1 NB: line looks like F E**Example 6**V is (-5, -9) and W is (12,-9). • mVW = (y2-y1) (x2-x1) • = (-9 -(- 9)) • ( 12 - (-5)) • = 0/17 • = 0 • NB: A horizontal line has no steepness. NB: line looks like V W**Example 7**S is (5, -9) and T is (5, 12). • mST = (y2-y1) (x2-x1) • = (12 -(- 9)) • ( 5 - 5) • = 21/0 • and this is undefined • or infinite NB: line looks like T Vertical line has infinite gradient. S**Summary of Gradients**Positive gradient goes uphill. Negative gradient goes downhill. Zero gradient is horizontal. Infinite gradient is vertical.**Parallel Lines**• Parallel lines run in the same direction so must be equally steep. • Hence parallel lines have equal gradients. • Example 8 • Prove that if A is (4,-3) , B is (9,3) C is (11,1) & D is (2, -1) • then ACBD is a parallelogram**Ex8 ctd**D B NB; The order of the letters is important. A • mAC = (1 + 3)/(11- 4) = 4/7 • mDB = (3 + 1)/(9 - 2) = 4/7 • mAC = mDB so AC is parallel to DB • mAD = (-1 + 3)/(2 - 4) = 2/-2 = -1 • mCB = (3 - 1)/(9 - 11) = 2/-2 = -1 • mAD = mCB so AC is parallel to DB • Since the opposite sides are parallel then it follows that ACBD is a parallelogram. C**COLLINEARITY**Defn: Three or more points are said to be collinear if the gradients from any one point to all the others is always the same. Example 8a K is (5, -8), L is (-2, 6) and M is (9, -16). Prove that the three points are collinear. 6 - (-8) 14 = -2 mKL = = Since KL & KM have equal gradients and a common point K then it follows that K, L & M are collinear. -2 - 5 -7 -16 - (-8) -8 mKM = = = -2 9 - 5 4**Ex8b**A Navy jet flies over two lighthouses with map coordinates (210,115) & (50,35). If it continues on the same path will it pass over a yacht at (10,15) ? m1 = (115-35)/(210-50) = 80/160 = 1/2 m2 = (115-15)/(210-10) = 100/200 = 1/2 Since gradients equal & (210,115) a common point then the three places are collinear so plane must fly over all three.