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High School Mathematics - 29.6 Derivation of distance, external and internal division formulae

 Distance Formula The distance between the points P(x1, y1) and Q(x2, y2) is given by the formula d =� [(x2-x1)2 + (y2-y1)2] Let P(x1, y1) and Q(x2, y2) be the two points. From P, Q draw PL, QM perpendiculars on the x-axis and PR perpendicular on QM. Then, PR = LM = x2 - x1 RQ = MQ - MR = MQ - LP = y2 - y1 From right angled triangle PRQ, PQ2 = PR2 + RQ2 d2 = (x2 - x1)2 + (y2 - y1)2 d = �2 - x1)2 + (y2 - y1)2] Distance from the origin The distance of the point (x1, y1) from the origin is d = �1-0)2 + (y1-0)2] d = �12 + y12] Example: Find the distance between the points (6, -8), (2, -5). d2 = (x2-x1)2 + (y2-y1)2 d2 = (2-6)2 + [-5-(-8)]2 d2 = (-4)2 + (3)2 d2 = 16 + 9 = 25 d = 5 Answer: 5 units Division or section formula To find the co-ordinates of the point which divides internally the line joining two given points in a given ratio. Let A(x1, y1) and B(x2, y2) be the two given points and P a point on AB which divides it in the given ratio m1: m2. It is required to find the co-ordinates of P. Suppose they are (x, y). Draw the perpendiculars AL, PM, BN on OX and Ak, PT on PM and BN respectively. Then, from similar triangles APK and PBT we have AP/PB = AK/PT = KP/TB - 1 Now AP: PB = m1 : m2, AK = LM = OM - OL = x - x1 PT = MN = MK = MP - LA = y - y1 TB = NB - NT = NB - MP = y2 - y From 1 we have m1/m2 = (x - x1)/(x2 - x) = (y - y1)/(y2 - y) The first two relations give m1/m2 = (x - x1)/(x2 - x) or m1x2 - m1x = m2x - m2x1 Thus x = (m1x2 + m2x1)/(m1 + m2) , y = (m1y2 + m2y1)/(m1 + m2) The co-ordinates of mid-point are [(x1+x2)/2, (y1+y2)/2] Formula for external division Construct as shown in figure. From similar triangles APK and PBT, AP/PB = AK/PT = KP/TB or m1/m2 = (x - x1)/(x - x2) = (y - y1)/(y - y2) From the above we get x = (m1x2 - m2x1)/(m1 - m2) y = (m1y2 - m2y1)/(m1 - m2) Example: Find the coordinates of the point which divides the join of the points (8,9) and (-7,4) a. Internally in the ratio 2:3 b. Externally in the ratio 4:3 Solution:a. The coordinates of the point in the first case x = (2 x (-7) + 3 x 8)/(2+3) = (-14 + 24)/5 = 2 y = 2x4+3x9/(2+3) = 35/5 = 7 (2,7) b. For external division the co-ordinates are x = 4x(-7)-3x8/*4-3) = -52 y = 4x4-3x9/(4-3) = -11 (-52,-11) Directions: Solve the following problems.
 Q 1: Find the coordinates of the point which divides externally the join of the points (3,4) and (-6,2) in the ratio 3:2.(21,8)(8,21)(-6,5) Q 2: Determine the ratio in which y-x+2 = 0 divides the line joining (3,-1) and (8,9).2:3 externally1:5 internally2:3 internally Q 3: In what ratio is the line joining the points (2,-3) and (5,6) divided by the x-axis.1:21:72:4 Q 4: Find the coordinates of the point which divides externally the join of the points (-4,4) and (1,7) in the ratio 2:1.(5,-6)(3,7)(6,10) Q 5: Find the coordinates of the point which divides internally the join of the points (8,9) and (7,-4) in the ratio 2:3.(-2,7)(2,7) or (1,6)(-1,-5) Q 6: In what ratio does the point (1,-7/2) divide the join of (-2,-4) and (2,-10/3).3:11:31:6 Q 7: Find the coordinates of the point which divides internally the join of the points (1,-2) and (4,7) in the ratio 1:2.(-5,-4)(-4,-7)(2,1) or (3,4) Q 8: Find the coordinates of the points of trisection of the line joining the points (2,3) and (6,5).(10/3, 11/3), (14/3, 13/3)(0,5), (1,5)(4,8), (7,0) Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only!