// Numbas version: finer_feedback_settings {"name": "Julie's copy of Q3 Given 2 lines, Coordinate Geometry 1", "extensions": ["jsxgraph"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"functions": {"inter": {"definition": "\n var a = Numbas.jme.unwrapValue(scope.variables.p_m);\n var b = Numbas.jme.unwrapValue(scope.variables.p_c);\n var c = Numbas.jme.unwrapValue(scope.variables.r_m);\n var d = Numbas.jme.unwrapValue(scope.variables.r_c);\n\n var miny = Numbas.jme.unwrapValue(scope.variables.miny);\n var maxy = Numbas.jme.unwrapValue(scope.variables.maxy);\n var minx = Numbas.jme.unwrapValue(scope.variables.minx);\n var maxx = Numbas.jme.unwrapValue(scope.variables.maxx);\n var div = Numbas.extensions.jsxgraph.makeBoard('600px','600px',\n {boundingBox:[minx,maxy,maxx,miny],\n axis:true,\n showNavigation:false,\n grid:true});\n var brd = div.board; \n var li1=brd.create('line',[[0,-b],[1,a-b]],{fixed:true,name:'Line P',withLabel:true});\n var li2=brd.create('line',[[0,d],[1,c+d]],{fixed:true,name:'Line R',withLabel:true});\n \n\n return div;\n ", "type": "html", "language": "javascript", "parameters": []}}, "ungrouped_variables": ["pr", "c1", "c2", "a", "b", "val1", "above1", "below1", "on1", "mark1a", "mark1b", "mark1o", "val2", "above2", "below2", "on2", "mark2a", "mark2b", "mark2o"], "name": "Julie's copy of Q3 Given 2 lines, Coordinate Geometry 1", "tags": ["rebelmaths"], "preamble": {"css": "", "js": ""}, "advice": "
P => 3x - 2y -6 = 0 and R => 2x + 3y -35 = 0
\na) point a(4,-2) on, above or below P
\nP => y = $\\frac{3}{2}$x - 3
\n$\\frac{3}{2}$(4) - 3 = 3
\n-2 < 3
\n=> point a is below P
\nb) point b(16,1) on, above or below R
\nR => y = -$\\frac{2}{3}$x + $\\frac{35}{3}$
\n-$\\frac{2}{3}$(16) + $\\frac{35}{3}$ = 1
\n1 = 1
\n=> point b is on R
\nc) Find the slope of P and the slope of R.
\nslope of P (m1) = $ \\frac{3}{2}$
\nslope of R (m2) = $ -\\frac{2}{3}$
\n$ \\frac{3}{2} \\times -\\frac{2}{3}$ = -1
\n=> P is perpendicular to R
\nd)
\n3x - 2y -6 = 0 (X 3) => 9x - 6y = 18
\n2x + 3y -35 = 0 (X 2) => 4x + 6y = 70
\n=> 13x = 88
\nx = 6.77
\n3(6.77) - 2y -6 = 0
\ny = 7.16
\npoint of intersection = (6.77 , 7.16)
\ne) x-axis intersect of P
\ny=0 => 3x - 2(0) -6 = 0 => x = 2
\npoint = (2 , 0)
\ny-axis intersect of P
\nx=0 => 3(0) - 2y -6 = 0 => y = -3
\npoint = (0 , -3)
\nf) $(\\frac{4+16}{2},\\frac{-2+1}{2}) = (10,-\\frac{1}{2})$
\ng) Using the formula dis = $\\sqrt((X2-X1)^2 + (Y2-Y1)^2)$, where (X1,Y1) = b(16,1) and (X2,Y2) = a(4,-2).
\n$\\sqrt((4-16)^2 + (-2-1)^2)$
\n$\\sqrt(144 + 9) = 12.37$
\nh) Point (a,2) is on R
\n2x + 3y -35 = 0
\n2(a) + 3(2) -35 = 0
\n2a = 29
\na = 14.5
\ni) Line parallel to R contains (-2,5) and (p,4)
\n$\\frac{4-5}{p-(-2)} = -\\frac{2}{3}$
\n$p = -\\frac{1}{2}$
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\nFind the slope of line P:
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\nFind the slope of line R:
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\nIs P perpendicular to R.
\n[[2]]
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\nGive answer in fraction form.
\nPoint c, R intersects x-axis = ([[0]],[[1]])
\nPoint d, R intersects y-axis = ([[2]],[[3]])
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\n([[0]],[[1]])
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\nans = [[0]]
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\nGive answer to 2 decimal places.
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\nP is the line $\\var{pr[0]}x - \\var{pr[1]}y - \\var{c1} = 0$ and R is the line $\\var{pr[1]}x + \\var{pr[0]}y - \\var{c2} = 0$.
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Graph shown
\nRebelmaths
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