Topic Coordinate Geometry Estimated reading: 6 minutes 113 views Equation of a LineThe General Equation of a Straight LineDerive the general equation of a straight lineCOORDINATES OF A POINT•The coordinates of a points – are the values of x and y enclosed by the brackets which are used to describe the position of point in a line in the plane.The plane is called xy-plane and it has two axis.horizontal axis known as axis andvertical axis known as axisConsider the xy-plane belowThe coordinates of points A, B, C ,D and E are A(2, 3), B(4, 4), C(-3, -1), D(2, -4) and E(1, 0).DefinitionExample 1Find the gradient of the lines joiningExample 2(a) The line joining (2, -3) and (k, 5) has a gradient -2. Find kExercise 11. Find the gradient of the line which passes through the following points ;(3,6) and (-2,8)(0,6) and (99,-12)(4,5)and (5,4)2. A line passes through (3, a) and (4, -2), what is the value of a if the slope of the line is 4?3. The gradient of the line which goes through (4,3) and (-5,k) is 2. Find the value of k.FINDING THE EQUATION OF A STRAIGHT LINEThe equation of a straight line can be determined if one of the following is given:-Example 3Find the equation of the line with the followingGradient 2 and interceptGradient and passing through the pointPassing through the points andSolutionEQUATION OF A STRAIGHT LINE IN DIFFERENT FORMSThe equation of a line can be expressed in two formsExample 4Find the gradient of the following linesINTERCEPTSThereforeExample 5Find the y-intercept of the following linesExample 6Find the x and y-intercept of the following linesExercise 2Attempt the following Questions.Find the y-intercept of the line 3x+2y = 18 .What is the x-intercept of the line passing through (3,3) and (-4,9)?Calculate the slope of the line given by the equation x-3y= 9Find the equation of the straight line with a slope -4 and passing through the point (0,0).Find the equation of the straight line with y-intercept 5 and passing through the point (-4,8).GRAPHS OF STRAIGHT LINESThe graph of straight line can be drawn by using the following methods;By using interceptsBy using the table of valuesExample 7Sketch the graph of Y = 2X – 1SOLVING SIMULTANEOUS EQUATION BY GRAPHICAL METHODUse the intercepts to plot the straight lines of the simultaneous equationsThe point where the two lines cross each other is the solution to the simultaneous equationsExample 8Solve the following simultaneous equations by graphical methodExercise 31. Draw the line 4x-2y=7 and 3x+y=7 on the same axis and hence determine their intersection point2. Find the solution for each pair the following simultaneous equations by graphical method;y-x = 3 and 2x+y = 93x- 4y=-1 and x+y = 2x = 8 and 2x-3y = 10Midpoint of a Line SegmentThe Coordinates of the Midpoint of a Line SegmentDetermine the coordinates of the midpoint of a line segmentLet S be a point with coordinates (x1,y1), T with coordinates (x2,y2) and M with coordinates (x,y) where M is the mid-point of ST. Consider the figure below:Considering the angles of the triangles SMC and TMD, the triangles SMC and TMD are similar since their equiangularExample 9Find the coordinates of the mid-point joining the points (-2,8) and (-4,-2)SolutionTherefore the coordinates of the midpoint of the line joining the points (-2,8) and (-4, -2) is (-3,3).Distance Between Two Points on a PlaneThe Distance Between Two Points on a PlaneCalculate the distance between two points on a planeConsider two points, A(x1,y1) and B(x2,y2) as shown in the figure below:The distance between A and B in terms of x1, y1,x2, and y2can be found as follows:Join AB and draw doted lines as shown in the figure above.Then, AC = x2– x1and BC = y2– y1Since the triangle ABC is a right angled, then by applying Pythagoras theorem to the triangle ABC we obtainTherefore the distance is 13 units.Parallel and Perpendicular LinesGradients in order to Determine the Conditions for any Two Lines to be ParallelCompute gradients in order to determine the conditions for any two lines to be parallelThe two lines which never meet when produced infinitely are called parallel lines. See figure below:The two parallel lines must have the same slope. That is, if M1is the slope for L1and M2is the slope for L2thenM1= M2Gradients in order to Determine the Conditions for any Two Lines to be PerpendicularCompute gradients in order to determine the conditions for any two lines to be perpendicularWhen two straight lines intersect at right angle, we say that the lines are perpendicular lines. See an illustration below.Consider the points P1(x1,y1), P2(x2,y2), P3(x3,y3), R(x1,y2) and Q(x3,y2) and the anglesα,β,γ(alpha, beta and gamma respectively).α+β = 90 (complementary angles)α+γ= 90 (complementary angles)β = γ (alternate interior angles)Therefore the triangle P2QP3is similar to triangle P1RP2.Generally two perpendicular lines L1and L2with slopes M1and M2respectively the product of their slopes is equal to negative one. That is M1M2= -1.Example 10Show that A(-3,1), B(1,2), C(0,-1) and D(-4,-2) are vertices of a parallelogram.SolutionLet us find the slope of the lines AB, DC, AD and BCWe see that each two opposite sides of the parallelogram have equal slope. This means that the two opposite sides are parallel to each other, which is the distinctive feature of the parallelogram. Therefore the given vertices are the vertices of a parallelogram.Problems on Parallel and Perpendicular LinesSolve problems on parallel and perpendicular linesExample 11Show that A(-3,2), B(5,6) and C(7,2) are vertices of a right angled triangle.SolutionRight angled triangle has two sides that are perpendicular, they form 90°.We know that the slope of the line is given by: slope = change in y/change in xNow,Since the slope of AB and BC are negative reciprocals, then the triangle ABC is a right angled triangle at B.Tagged:form 4MathematicsNotes Next - Topic Area And Perimeter