 Hello and welcome to the session. In this session we are going to discuss concurrency of three lines. The three given lines are concurrent if they meet in a point that is they have a point of intersection to prove that the three given lines are concurrent. We follow the following steps. The first step is solve the equations of two lines and we coordinate the point of intersection. We show that these coordinates satisfy the third equation 1x plus b1y plus c1 is equal to 0 and not this equation as 1 a2x plus b2y plus c2 is equal to 0 and not this equation as 2b3y plus c3 is equal to 0 and not this equation as 3b3 concurrent lines. Then the point of intersection of equation 1 and equation 2 must lie on the third that is the point of intersection of equation 1 and equation 2 is b1c2 minus b2c1 upon a1b2 minus a2b1 c1a2 minus a1c2 whole upon a1b2 minus a2b1 and this point of intersection must lie on equation 3. Therefore we have a3x that is b1c2 minus b2c1 whole upon a1b2 minus a2b1 plus b3y that is c1a2 minus a1c2 whole upon a1b2 minus a2b1 plus c3 is equal to 0. On solving this equation we get a3 into b1c2 minus b2c1 plus b3 into c1a2 minus a1c2 plus c3 into a1b2 minus a2b1 is equal to 0 which can be written in the determinant form as the determinant containing elements a1b1c1 a2b2 c2 a3b3c3 is equal to 0 and this is the required condition for the concurrency of three lines. Now we are going to discuss the test for concurrency of three straight lines. The three lines that is l1 which is equivalent to a1x plus b1y plus c1 is equal to 0 l2 which is equivalent to a2x plus b2y plus c3 is equal to 0 and l3 which is equivalent to a3x plus b3y plus c3 is equal to 0 are concurrent. It is possible to find three constants lambda1 lambda2 lambda3 other than 0 such that lambda1 into l1 plus lambda2 into l2 plus lambda3 into l3 is equal to 0 that is lambda1 into l1 which is equivalent to a1x plus b1y plus c1 plus lambda2 into l2 that is a2x plus b2y plus c2 plus lambda3 into l3 which is equivalent to a3x plus b3y plus c3 is equal to 0 Now let x1 y1 be the point of intersection of l1 and l2 then a1 into x1 plus b1 into y1 plus c1 is equal to 0 and lambda's equation has 1 2 into x1 plus b2 into y1 plus c2 is equal to 0 lambda's equation has 2 as we know that lambda1 into l1 plus lambda2 into l2 plus lambda3 into l3 is equal to 0 is an identity therefore it is true for all values of x and y and here it is true for x is equal to x1 and y is equal to y1 therefore the value of lambda1 into a1x1 plus b1y1 plus c1 plus lambda2 into a2x1 plus b2y1 plus c2 plus lambda3 into a3x1 plus b3y1 plus c3 is equal to 0 from equation 1 and equation 2 we know that a1x1 plus b1y1 plus c1 is equal to 0 and a2x1 plus b2y1 plus c2 is equal to 0 therefore we have lambda1 into 0 plus lambda2 into 0 plus lambda3 into a3x1 plus b3y1 plus c3 is equal to 0 that is lambda3 into a3x1 plus b3y1 plus c3 is equal to 0 which implies that a3x1 plus b3y1 plus c3 is equal to 0 as it is given that lambda3 cannot be 0 therefore a3x plus b3y1 plus c3 is equal to 0 also passes through the point x1y1 hence the lines are component now we are going to learn about angular bisectors an angular bisector of two lines is defined as the locus of a point which is equidistant from the two lines if line L1 is given by plus b1y plus c1 is equal to 0 and line L2 which is given by a2x plus b2y plus c2 is equal to 0 r equations of the two lines hk be any point on the bisector L1 and L2 are the two lines and point p with the coordinates hk is any point on the bisector then we have modulus of p2 is equal to modulus of pr where p2 is perpendicular to L1 and pr is perpendicular to L2 so we have modulus of pq is equal to modulus of pr where pq is the perpendicular distance of point p from line L1 and modulus of pq is equal to modulus of a1h plus b1k plus c1 upon square root of a1 square plus b1 square similarly pr is the perpendicular distance of point p from line L2 and modulus of pr is given by modulus of a2h plus b2k plus c2 upon square root of a2 square plus b2 square as we know modulus of pq is equal to modulus of pr therefore we have modulus of a1h plus b1k plus c1 upon square root of a1 square plus b1 square is equal to modulus of a2h plus b2k plus c2 upon square root of a2 square plus b2 square therefore locus of point p which coordinates hk is given by modulus of a1x plus b1y plus c1 upon square root of a1 square plus b1 square is equal to modulus of a2h plus b2y plus c2 upon square root of a2 square plus b2 square the equations by vectors of the angle between lines l1 and l2 is given by and y plus c1 upon square root of a1 square plus b1 square is equal to plus minus of a2x plus b2y plus c2 upon square root of a2 square plus b2 square on taking the positive sign we get the equation of the bisector of acute angle between the lines on taking the negative sign we get the equation of the bisector of the obtuse angle the bisector can be distinguished with respect to the original scholars first is make the constant terms positive in the equations of the given lines next we have the equation obtained on taking the positive sign will be the equation of the bisector b1 that bisects the angle in which the origin lies bisector b1 bisects the angle in which the origin lies the equation obtained in taking the negative sign equation of the bisector b2 bisects the angle in which does not lie bisector b2 bisects the angle in which the origin does not lie this completes our lesson hope you enjoyed this lesson