 Hello and welcome to the session. In this session, first we are going to discuss about locus and its equation. Let us discuss the definition of locus. If a point obeys some geometrical conditions, then the path so traced is known as the locus of that point. Or in other words, we can say that locus of a point is the path which it describes when moving under a certain given condition or conditions. For example, we have the locus of a point which moves in a plane keeping a constant distance from a fixed given point is a circle. We have another example, the locus of a point which moves in a plane keeping a constant distance from the fixed line is another straight line parallel to the given line. Next we are going to discuss the method to find the equation of the locus by the given condition. If a point moves, then there is some algebraic relation between the coordinates of that moving point. Such a relation is called the equation of the locus. To find the equation of the locus, we follow some steps. The first step is take a moving point as p with the coordinates x, y on the locus. Next, find the condition that the moving point obeys. Next, express the given condition with the help of x and y and then simplify. Then this is the required equation of the locus. The coordinates of the moving point p are known as current coordinates. We can also follow the following steps. First, take any point with the coordinates h, k on the locus. Next, use the given condition and write a relation between h and k. Then simplify the relation if needed. Then change into current coordinates x, y that does put the value of x equal to h and y equal to k. The equation so obtained if we require the equation of the locus, let us take an example. Find the locus of a point which moves so that its distance from x axis is twice its distance from the y axis. Here, let h, k be any point on the locus. Then by given condition modulus of h is equal to twice of modulus of k that is modulus of x is equal to twice of modulus of y as we have put the value of h of x and k as y. Therefore, the locus of is given by x is equal to plus minus 2y. Now we shall discuss how to reduce the general equation ax plus by plus c is equal to 0 to the prependicular form. Let x cos alpha plus y sin alpha is equal to p mark this equation as 1 db normal form ax plus by plus c is equal to 0 mark this equation as 2. Then equation 1 and equation 2 represent the same straight line. Therefore, a upon cos alpha is equal to b upon sin alpha is equal to c upon minus p. This implies that cos of alpha is equal to minus ap upon cos alpha is equal to minus bp upon c. This implies that cos square alpha plus sin square alpha is equal to a square p square upon c square plus sin square alpha is minus of bp upon c. Therefore, sin square alpha will be b square p square upon c square which implies that cos square alpha plus sin square alpha which is equal to 1 is equal to a square p square upon c square plus b square p square upon c square which implies that 1 is equal to taking p square upon c square common into a square plus b square which further implies that c square is equal to p square into a square plus b square. Or we can also write p square is equal to c square upon a square plus b square which implies that p is equal to plus minus c upon square root of a square plus b square p denotes the length of the perpendicular from the origin to the line and is always positive. Therefore, p is equal to c upon square root of a square plus b square. Now put the value of p that is c upon square root of a square plus b square in cos of alpha is equal to minus a into p upon c and sin of alpha which is equal to minus of b into p upon c. Therefore, cos of alpha is equal to minus of a into c upon c into square root of a square plus b square which is equal to minus of a upon square root of a square plus b square of alpha is equal to minus of b into c upon c into square root of a square plus b square which is equal to minus b upon square root of a square plus b square. Therefore, equation x cos of alpha plus y sin of alpha minus p is equal to 0 becomes minus a upon square root of a square plus b square into x minus b upon square root of a square plus b square into y minus c upon square root of a square plus b square is equal to 0 which can also be written as minus a upon square root of a square plus b square into x minus b upon square root of a square plus b square into y is equal to c upon square root of a square plus b square which is the required normal form of the line b y plus c is equal to 0. Therefore, the working rule to obtain the equation of the line in normal form is given by divide throughout by square root of a square plus b square and then transpose the constant term to the right hand side Take it positive by changing the sign if necessary. Now we shall discuss the property of the identical lines if two equations that is a x plus b y plus c is equal to 0 and a dash x plus b dash y plus c dash is equal to 0 represent the same straight line then the ratio of the coefficients of x that is a upon a dash is equal to the ratio of the coefficients of y that is b upon b dash is equal to the ratio of the constant terms that is c upon c dash Now as the two equations represent the same straight line so they will have the same slope and the same y intercepts we mark the equation a x plus b y plus c is equal to 0 as 1 and a dash x plus b dash y plus c dash is equal to 0 as 2 Now let us represent the equation 1 and equation 2 in the form of y is equal to mx plus c then we have equation 1 becomes y is equal to minus a upon b into x minus c upon b and equation 2 becomes y is equal to minus a dash upon b dash into x minus c dash Now as slopes being equal the slope of equation 1 is minus a upon b and the slope of equation 2 is minus a dash upon b dash therefore minus a upon b is equal to minus a dash upon b dash which implies that a upon a dash is equal to b upon b dash Now this equation as 3 y intercepts are also equal therefore minus c upon b is equal to minus c dash upon b dash which implies that c upon c dash is equal to b upon b dash Now this equation as 4 now from equation 3 and equation 4 we say that a upon a dash is equal to b upon b dash is equal to c upon c dash Now let us discuss intersection of straight lines let equations of the two lines be o1x plus b1y plus c1 is equal to 0 mark this equation as 1 and a2x plus b2y plus c2 is equal to 0 mark this equation as 2 We know that the coordinates of the point or points of intersection of the two lines satisfy the equation of both of them and so their coordinates are obtained by solving their equations simultaneously And the corresponding values of x and y so obtained give the value of abscissa and ordinate of the point of intersection Now suppose these two lines intersect at the point p with the coordinates x1, y1 then x1, y1 satisfies each of the given equation Therefore a1x1 plus b1y1 plus c1 is equal to 0 and a2x1 plus b2y1 plus c2 is equal to 0 Now on solving these equations by cross multiplication we get x1 upon b1c2 minus b2c1 is equal to minus of y1 upon a1c2 minus a2c1 is equal to 1 upon a1v2 minus a2b1 which implies that x1 is equal to b1c2 minus b2c1 upon a1b2 minus a2b1 and y1 is equal to a2c1 minus a1c2 upon a1b2 minus a2b1 Therefore the coordinates of the point of intersection of equations 1 and 2 b1c2 minus b2c1 upon a1b2 minus a2b1 a2c1 minus a1c2 upon a1b2 minus a2b1 This completes our session. Hope you enjoyed this session.