 Hello and welcome to the session. In this session, let us first discuss family of lines, a set of lines, satisfying a given condition a family of lines. A family of lines can be represented by a linear equation in x and y involving one ordinary constant which is called parameter of the family of lines under consideration. We come across various forms of equations of lines as y is equal to mx plus c or x cos lambda plus y sin lambda is equal to p or x upon a plus y upon b is equal to 1. Each of these equations has two constants which have some geometrical significance which implies that when some definite values are assigned to these constants we obtain a particular equation of the line. Let us take an example for the equation y is equal to mx plus c the constants m and c are fixed for a particular line but vary from line to line and these are called parameters. For the equation y is equal to mx plus c m and c are the parameters for the equation x cos lambda plus y sin lambda is equal to p lambda and p are the parameters and for the equation x upon a plus y upon b is equal to 1. a and b are the parameters if we fix one of the parameter and assign different values to the other parameters then we will obtain different lines for different values of this variable parameter. All these lines will have a common property are called as family of lines or one parameter system. Therefore the equation y is equal to x plus c in which c is a parameter represents the one parameter family lines with slope equal to 1 as the value of theta is equal to 45 degrees and tan of 45 degrees is equal to 1 therefore slope is equal to 1 and there is an infinite number of lines in the family similarly the equation y is equal to mx plus 5 where m is the parameter represents an infinite number of lines whose y intercept is 5. Now we are going to discuss about family of lines parallel to the given line. Let the given equation of the line be ax plus by plus c is equal to 0. Now we need to find the family of lines parallel to this given line and the required equation will be ax plus by plus k is equal to 0 for different values of k we can have family of lines parallel to the given equation of line here we change the constant term with the new constant term and the remaining terms will be as it is. Now we are going to learn about family of lines perpendicular to the given line. Let the given equation of the line be ax plus by plus c is equal to 0. Now we need to find family of lines perpendicular to this line. We know that if two lines are perpendicular then their slopes are negative reciprocal to each other. So here we interchange the coefficients of x and y in the given equation of line. Now change the sign of either of x or y and then replace the constant term by a new constant k therefore the required equation will be minus of ay plus bx plus k is equal to 0 for different values of k it represents a family of lines perpendicular to the given line to obtain the family of lines perpendicular to the given line ax plus by plus c we first interchange the coefficients of x and y in the given equation of line then we change the sign of either of x or y and replace the constant term by a new constant say k and thus we get the required equation of the family of lines perpendicular to the given lines. Now we are going to learn about how to find the local of the point of intersection of two lines whose equation involves one parameter. Here we eliminate that one parameter from the equation of the lines and the element is the required locus. Let us take an example find the locus of the point of intersection of the lines x plus y is equal to 5 minus lambda and towards minus 3 y is equal to 2 plus 4 lambda when lambda is a variable we have given the equation of the lines as 2 plus y is equal to 5 minus lambda mark it as equation 1 and towards minus 3 y is equal to 2 plus 4 lambda mark this equation as 2 the required equation of the locus will be obtained by eliminating lambda between the lines from equation 1 we get lambda is equal to 5 minus x minus y now substitute the value of lambda in equation 2 we get 2 x minus 3 y is equal to 2 plus 4 into lambda that is 5 minus x minus y which implies that 2 x minus 3 y is equal to 2 plus 4 into 5 that is 20 minus 4 into x that is 4 x minus 4 into y that is 4 y which further implies that 2 x plus 4 into x minus 3 y plus 4 y minus 22 is equal to 0 that is plus y minus 22 is equal to 0 which is the required equation of the locus this completes our session but you enjoyed the lesson.