<span>Vector Equation (Line)</span>(x,y) = (x,y) + t(a,b);tERParametric Formx = x + t(a), y = y + t(b); tERr = (-4,-2) + t((-3,5);tERFind the vector equation of the line passing through A(-4,-2) & parallel to m = (-3,5)<span>Point: (2,5) Create a direction vector: AB = (-1 - 2, 4 - 5) = (-3,-1) or (3,1)when -1 (or any scalar multiple) is divided out. r = (2,5) + t(-3,-1);tER</span>Find the vector equation of the line passing through A(2,5) & B(-1,4)<span>x = 4 - 3t y = -2 + 5t ;tER</span>Write the parametric equations of the line passing through the line passing through the point A(4,-2) & with a direction vector of m =(-3,5)<span>Create Vector Equation first: AB = (2,8) Point: (4,-3) r = (4,-3) + (2,8); tER x = 4 + 2t y = -3 + 8t ;tER</span>Write the parametric equations of the line through A(4,-3) & B(6,5)<span>Make parametric equations: x = 5 + 4t y = -2 + 3t ; tER For x sub in -3 -3 = 5 + 4t (-8 - 5)/4 = t -2 = t For y sub in -8 -8 = -2 + 3t (-8 + 2)/3 = t -2 = t Parameter 't' is consistent so pt(-3,-8) is on the line.</span>Given the equation r = (5,-2) + t(4,3);tER, is (-3,-8) on the line?<span>Make parametric equations: x = 5 + 4t y = -2 + 3t ; tER For x sub in 1 -1 = 5 + 4t (-1 - 5)/4 = t -1 = t For y sub in -7 -7 = -2 + 3t (-7 + 2)/3 = t -5/3 = t Parameter 't' is inconsistent so pt(1,-7) is not on the line.</span>Given the equation r = (5,-2) + t(4,3);tER, is (1,-7) on the line?<span>Use parametric equations when generating points: x = 5 + 4t y = -2 + 3t ;tER X-int: sub in y = 0 0 = -2 + 3t solve for t 2/3 = t (this is the parameter that will generate the x-int) Sub t = 2/3 into x = 5 + 4t x = 5 + 4(2/3) x = 5 + (8/3) x = 15/3 + (8/3) x = 23/3 The x-int is (23/3, 0)</span>What is the x-int of the line r = (5,-2) + t(4,3); tER?Note: if they define the same line: 1) Are their direction vectors scalar multiples? 2) Check the point of one equation in the other equation (LS = RS if point is subbed in)What are the two requirements for 2 lines to define the same line?
