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Page 1 JEE Solved Example on Straight Lines JEE Mains Q1: The opposite angular points of a square are ( ?? , ?? ) and ( ?? , - ?? ) . Then the co-ordinates of other two points are (A) D( ?? ?? , ?? ?? ) , ?? ( - ?? ?? , ?? ?? ) (B) ?? ( ?? ?? , ?? ?? ) , ?? ( ?? ?? , ?? ?? ) (C) ?? ( ?? ?? , ?? ?? ) , ?? ( - ?? ?? , ?? ?? ) (D) None of these Ans: (C) Obviously, slope of AC = 5 / 2. Let ?? be the slope of a line inclined at an angle of 45 ° to ???? , then tan ? 45 ° = ± ?? - 5 2 1 + ?? · 5 2 ? ?? = - 7 3 · 3 7 . Thus, let the slope of ???? or ???? be 3/7and that of ???? or ???? be - 7 3 . Then equation of ???? is 3x - 7y + 19 = 0. Also the equation of ???? is 7x + 3y - 4 = 0. On solving these equations, we get, B ( - 1 2 , 5 2 ). Now let the coordinates of the vertex ?? be ( h , ?? ) . Since the middle points of ???? and ???? are same, therefore 1 2 ( ?? - 1 2 ) = 1 2 ( 3 + 1 ) ? ?? = 9 2 , 1 2 ( ?? + 5 2 ) = 1 2 ( 4 - 1 ) ? ? ?? = 1 2 . Hence, ?? = ( 9 2 , 1 2 ) . Q2: If the lines ?? = ?? ?? + ?? and ?? ?? = ?? + ?? are equally inclined to the line ?? = ???? + ?? , then ?? = (A) ?? + ?? v ?? ?? (B) ?? - ?? v ?? ?? (C) ?? ± ?? v ?? ?? (D) ?? ± ?? v ?? ?? Ans: (D) ?? 1 = 3 , ?? 2 = 1 2 and ?? 3 = ?? Let the angle between first and third line is ?? , and between second and third is ?? 2 , then Page 2 JEE Solved Example on Straight Lines JEE Mains Q1: The opposite angular points of a square are ( ?? , ?? ) and ( ?? , - ?? ) . Then the co-ordinates of other two points are (A) D( ?? ?? , ?? ?? ) , ?? ( - ?? ?? , ?? ?? ) (B) ?? ( ?? ?? , ?? ?? ) , ?? ( ?? ?? , ?? ?? ) (C) ?? ( ?? ?? , ?? ?? ) , ?? ( - ?? ?? , ?? ?? ) (D) None of these Ans: (C) Obviously, slope of AC = 5 / 2. Let ?? be the slope of a line inclined at an angle of 45 ° to ???? , then tan ? 45 ° = ± ?? - 5 2 1 + ?? · 5 2 ? ?? = - 7 3 · 3 7 . Thus, let the slope of ???? or ???? be 3/7and that of ???? or ???? be - 7 3 . Then equation of ???? is 3x - 7y + 19 = 0. Also the equation of ???? is 7x + 3y - 4 = 0. On solving these equations, we get, B ( - 1 2 , 5 2 ). Now let the coordinates of the vertex ?? be ( h , ?? ) . Since the middle points of ???? and ???? are same, therefore 1 2 ( ?? - 1 2 ) = 1 2 ( 3 + 1 ) ? ?? = 9 2 , 1 2 ( ?? + 5 2 ) = 1 2 ( 4 - 1 ) ? ? ?? = 1 2 . Hence, ?? = ( 9 2 , 1 2 ) . Q2: If the lines ?? = ?? ?? + ?? and ?? ?? = ?? + ?? are equally inclined to the line ?? = ???? + ?? , then ?? = (A) ?? + ?? v ?? ?? (B) ?? - ?? v ?? ?? (C) ?? ± ?? v ?? ?? (D) ?? ± ?? v ?? ?? Ans: (D) ?? 1 = 3 , ?? 2 = 1 2 and ?? 3 = ?? Let the angle between first and third line is ?? , and between second and third is ?? 2 , then tan ? ?? 1 = 3 - ?? 1 + 3 ?? and tan ? ?? 2 = ?? - 1 2 1 + ?? 2 But ?? 1 = ?? 2 ? ? ? 3 - ?? 1 + 3 ?? = ?? - 1 2 1 + ?? 2 ? ? 7 ?? 2 - 2 ?? - 7 = 0 ? ?? = 1 ± 5 v 2 7 Q3: Let ?? ( - ?? , ?? ) , ?? ( ?? , ?? ) and ?? ( ?? , ?? v ?? ) be three points. Then the equation of the bisector of the angle ???? ?? is (A) v ?? ?? ?? + ?? = ?? (B) ?? + v ?? ?? = ?? (C) v ?? ?? + ?? = ?? (D) ?? + v ?? ?? ?? = ?? Ans: (C) Slope of QR = 3 v 3 - 0 3 - 0 = v 3 i.e., ?? = 60 ° Clearly, ? ?????? = 120 ° ???? is the angle bisector of the angle, so line ???? makes 120 ° with the positive direction of ?? -axis. Therefore, equation of the bisector of ? ?????? is ?? = tan ? 120 ° ?? or ?? = v 3 ?? i.e., v 3 ?? + ?? = 0. Q4: If for a variable line ?? ?? + ?? ?? = ?? , the condition ?? ?? ?? + ?? ?? ?? = ?? ?? ?? ? ( ?? is a constant) is satisfied, then locus of foot of perpendicular drawn from origin to the line is (A) ?? ?? + ?? ?? = ?? ?? / ?? (B) ?? ?? + ?? ?? = ?? ?? ?? (C) ?? ?? + ?? ?? = ?? ?? (D) ?? ?? - ?? ?? = ?? ?? Ans: (C) Equation of perpendicular drawn from origin to the line ?? ?? + ?? ?? = 1 is ?? - 0 = ?? ?? ( ?? - 0 ) [ ? m of given line = - ?? ?? , ? m of perpendicular = ?? ?? ] Page 3 JEE Solved Example on Straight Lines JEE Mains Q1: The opposite angular points of a square are ( ?? , ?? ) and ( ?? , - ?? ) . Then the co-ordinates of other two points are (A) D( ?? ?? , ?? ?? ) , ?? ( - ?? ?? , ?? ?? ) (B) ?? ( ?? ?? , ?? ?? ) , ?? ( ?? ?? , ?? ?? ) (C) ?? ( ?? ?? , ?? ?? ) , ?? ( - ?? ?? , ?? ?? ) (D) None of these Ans: (C) Obviously, slope of AC = 5 / 2. Let ?? be the slope of a line inclined at an angle of 45 ° to ???? , then tan ? 45 ° = ± ?? - 5 2 1 + ?? · 5 2 ? ?? = - 7 3 · 3 7 . Thus, let the slope of ???? or ???? be 3/7and that of ???? or ???? be - 7 3 . Then equation of ???? is 3x - 7y + 19 = 0. Also the equation of ???? is 7x + 3y - 4 = 0. On solving these equations, we get, B ( - 1 2 , 5 2 ). Now let the coordinates of the vertex ?? be ( h , ?? ) . Since the middle points of ???? and ???? are same, therefore 1 2 ( ?? - 1 2 ) = 1 2 ( 3 + 1 ) ? ?? = 9 2 , 1 2 ( ?? + 5 2 ) = 1 2 ( 4 - 1 ) ? ? ?? = 1 2 . Hence, ?? = ( 9 2 , 1 2 ) . Q2: If the lines ?? = ?? ?? + ?? and ?? ?? = ?? + ?? are equally inclined to the line ?? = ???? + ?? , then ?? = (A) ?? + ?? v ?? ?? (B) ?? - ?? v ?? ?? (C) ?? ± ?? v ?? ?? (D) ?? ± ?? v ?? ?? Ans: (D) ?? 1 = 3 , ?? 2 = 1 2 and ?? 3 = ?? Let the angle between first and third line is ?? , and between second and third is ?? 2 , then tan ? ?? 1 = 3 - ?? 1 + 3 ?? and tan ? ?? 2 = ?? - 1 2 1 + ?? 2 But ?? 1 = ?? 2 ? ? ? 3 - ?? 1 + 3 ?? = ?? - 1 2 1 + ?? 2 ? ? 7 ?? 2 - 2 ?? - 7 = 0 ? ?? = 1 ± 5 v 2 7 Q3: Let ?? ( - ?? , ?? ) , ?? ( ?? , ?? ) and ?? ( ?? , ?? v ?? ) be three points. Then the equation of the bisector of the angle ???? ?? is (A) v ?? ?? ?? + ?? = ?? (B) ?? + v ?? ?? = ?? (C) v ?? ?? + ?? = ?? (D) ?? + v ?? ?? ?? = ?? Ans: (C) Slope of QR = 3 v 3 - 0 3 - 0 = v 3 i.e., ?? = 60 ° Clearly, ? ?????? = 120 ° ???? is the angle bisector of the angle, so line ???? makes 120 ° with the positive direction of ?? -axis. Therefore, equation of the bisector of ? ?????? is ?? = tan ? 120 ° ?? or ?? = v 3 ?? i.e., v 3 ?? + ?? = 0. Q4: If for a variable line ?? ?? + ?? ?? = ?? , the condition ?? ?? ?? + ?? ?? ?? = ?? ?? ?? ? ( ?? is a constant) is satisfied, then locus of foot of perpendicular drawn from origin to the line is (A) ?? ?? + ?? ?? = ?? ?? / ?? (B) ?? ?? + ?? ?? = ?? ?? ?? (C) ?? ?? + ?? ?? = ?? ?? (D) ?? ?? - ?? ?? = ?? ?? Ans: (C) Equation of perpendicular drawn from origin to the line ?? ?? + ?? ?? = 1 is ?? - 0 = ?? ?? ( ?? - 0 ) [ ? m of given line = - ?? ?? , ? m of perpendicular = ?? ?? ] ? by - ax = 0 ? ?? ?? - ?? ?? = 0 Now, the locus of foot of perpendicular is the intersection point of line ?? ?? + ?? ?? = 1. and ?? ?? - ?? ?? = 0. To find locus, squaring and adding (i) and (ii) ( ?? ?? + ?? ?? ) 2 + ( ?? ?? - ?? ?? ) 2 = 1 ? ?? 2 ( 1 ?? 2 + 1 ?? 2 ) + ?? 2 ( 1 ?? 2 + 1 ?? 2 ) = 1 ? ?? 2 ( 1 ?? 2 ) + ?? 2 ( 1 ?? 2 ) = 1 , [ ? 1 ?? 2 + 1 ?? 2 = 1 ?? 2 ] ? ?? 2 + ?? 2 = ?? 2 . Q5: If the circle ?? ?? + ?? ?? - ?? ?? - ?? ?? + ?? = ?? does not touch or intersect the axes and the point ( ?? , ?? ) lies inside the circle, then (A) ?? < ?? < ?? (B) ?? < ?? < ???? (C) ?? < ?? < ???? (D) none of these Ans: (C) The centre of the given circle is ( 2 , 3 ) and the radius = v 4 + 9 - ?? i.e. v 13 - k. Since the given circle does not touch or intersect the coordinate axes and the point ( 2 , 2 ) lies inside the circle ? ?? -cooridnate of centre > radius i.e. 2 > v 13 - k, ?? -coordinate of centre > radius i.e. 3 > v 13 - k and 4 + 4 - 8 - 12 + ?? < 0 ? 4 > 13 - ?? , 9 > 13 - ?? and - 12 + ?? < 0 ? ?? > 9 , ?? > 4 and ?? < 12 ? 9 < ?? < 12 Q6: If ?? ?? , ?? ?? be the inclination of tangents drawn from the point ?? to the circle ?? ?? + ?? ?? = ?? ?? with ?? -axis, then the locus of ?? , it is given that ?? ???? ? ?? ?? + ?? ???? ? ?? ?? = ?? , is (A) ?? ( ?? ?? - ?? ?? ) = ?? ???? (B) ?? ( ?? ?? - ?? ?? ) = ?? ?? - ?? ?? (C) ?? ( ?? ?? - ?? ?? ) = ?? ???? (D) none of these Page 4 JEE Solved Example on Straight Lines JEE Mains Q1: The opposite angular points of a square are ( ?? , ?? ) and ( ?? , - ?? ) . Then the co-ordinates of other two points are (A) D( ?? ?? , ?? ?? ) , ?? ( - ?? ?? , ?? ?? ) (B) ?? ( ?? ?? , ?? ?? ) , ?? ( ?? ?? , ?? ?? ) (C) ?? ( ?? ?? , ?? ?? ) , ?? ( - ?? ?? , ?? ?? ) (D) None of these Ans: (C) Obviously, slope of AC = 5 / 2. Let ?? be the slope of a line inclined at an angle of 45 ° to ???? , then tan ? 45 ° = ± ?? - 5 2 1 + ?? · 5 2 ? ?? = - 7 3 · 3 7 . Thus, let the slope of ???? or ???? be 3/7and that of ???? or ???? be - 7 3 . Then equation of ???? is 3x - 7y + 19 = 0. Also the equation of ???? is 7x + 3y - 4 = 0. On solving these equations, we get, B ( - 1 2 , 5 2 ). Now let the coordinates of the vertex ?? be ( h , ?? ) . Since the middle points of ???? and ???? are same, therefore 1 2 ( ?? - 1 2 ) = 1 2 ( 3 + 1 ) ? ?? = 9 2 , 1 2 ( ?? + 5 2 ) = 1 2 ( 4 - 1 ) ? ? ?? = 1 2 . Hence, ?? = ( 9 2 , 1 2 ) . Q2: If the lines ?? = ?? ?? + ?? and ?? ?? = ?? + ?? are equally inclined to the line ?? = ???? + ?? , then ?? = (A) ?? + ?? v ?? ?? (B) ?? - ?? v ?? ?? (C) ?? ± ?? v ?? ?? (D) ?? ± ?? v ?? ?? Ans: (D) ?? 1 = 3 , ?? 2 = 1 2 and ?? 3 = ?? Let the angle between first and third line is ?? , and between second and third is ?? 2 , then tan ? ?? 1 = 3 - ?? 1 + 3 ?? and tan ? ?? 2 = ?? - 1 2 1 + ?? 2 But ?? 1 = ?? 2 ? ? ? 3 - ?? 1 + 3 ?? = ?? - 1 2 1 + ?? 2 ? ? 7 ?? 2 - 2 ?? - 7 = 0 ? ?? = 1 ± 5 v 2 7 Q3: Let ?? ( - ?? , ?? ) , ?? ( ?? , ?? ) and ?? ( ?? , ?? v ?? ) be three points. Then the equation of the bisector of the angle ???? ?? is (A) v ?? ?? ?? + ?? = ?? (B) ?? + v ?? ?? = ?? (C) v ?? ?? + ?? = ?? (D) ?? + v ?? ?? ?? = ?? Ans: (C) Slope of QR = 3 v 3 - 0 3 - 0 = v 3 i.e., ?? = 60 ° Clearly, ? ?????? = 120 ° ???? is the angle bisector of the angle, so line ???? makes 120 ° with the positive direction of ?? -axis. Therefore, equation of the bisector of ? ?????? is ?? = tan ? 120 ° ?? or ?? = v 3 ?? i.e., v 3 ?? + ?? = 0. Q4: If for a variable line ?? ?? + ?? ?? = ?? , the condition ?? ?? ?? + ?? ?? ?? = ?? ?? ?? ? ( ?? is a constant) is satisfied, then locus of foot of perpendicular drawn from origin to the line is (A) ?? ?? + ?? ?? = ?? ?? / ?? (B) ?? ?? + ?? ?? = ?? ?? ?? (C) ?? ?? + ?? ?? = ?? ?? (D) ?? ?? - ?? ?? = ?? ?? Ans: (C) Equation of perpendicular drawn from origin to the line ?? ?? + ?? ?? = 1 is ?? - 0 = ?? ?? ( ?? - 0 ) [ ? m of given line = - ?? ?? , ? m of perpendicular = ?? ?? ] ? by - ax = 0 ? ?? ?? - ?? ?? = 0 Now, the locus of foot of perpendicular is the intersection point of line ?? ?? + ?? ?? = 1. and ?? ?? - ?? ?? = 0. To find locus, squaring and adding (i) and (ii) ( ?? ?? + ?? ?? ) 2 + ( ?? ?? - ?? ?? ) 2 = 1 ? ?? 2 ( 1 ?? 2 + 1 ?? 2 ) + ?? 2 ( 1 ?? 2 + 1 ?? 2 ) = 1 ? ?? 2 ( 1 ?? 2 ) + ?? 2 ( 1 ?? 2 ) = 1 , [ ? 1 ?? 2 + 1 ?? 2 = 1 ?? 2 ] ? ?? 2 + ?? 2 = ?? 2 . Q5: If the circle ?? ?? + ?? ?? - ?? ?? - ?? ?? + ?? = ?? does not touch or intersect the axes and the point ( ?? , ?? ) lies inside the circle, then (A) ?? < ?? < ?? (B) ?? < ?? < ???? (C) ?? < ?? < ???? (D) none of these Ans: (C) The centre of the given circle is ( 2 , 3 ) and the radius = v 4 + 9 - ?? i.e. v 13 - k. Since the given circle does not touch or intersect the coordinate axes and the point ( 2 , 2 ) lies inside the circle ? ?? -cooridnate of centre > radius i.e. 2 > v 13 - k, ?? -coordinate of centre > radius i.e. 3 > v 13 - k and 4 + 4 - 8 - 12 + ?? < 0 ? 4 > 13 - ?? , 9 > 13 - ?? and - 12 + ?? < 0 ? ?? > 9 , ?? > 4 and ?? < 12 ? 9 < ?? < 12 Q6: If ?? ?? , ?? ?? be the inclination of tangents drawn from the point ?? to the circle ?? ?? + ?? ?? = ?? ?? with ?? -axis, then the locus of ?? , it is given that ?? ???? ? ?? ?? + ?? ???? ? ?? ?? = ?? , is (A) ?? ( ?? ?? - ?? ?? ) = ?? ???? (B) ?? ( ?? ?? - ?? ?? ) = ?? ?? - ?? ?? (C) ?? ( ?? ?? - ?? ?? ) = ?? ???? (D) none of these Ans: (C) Any tangent is ?? = ???? ± ?? v 1 + ?? 2 If it passes through ?? ( h , ?? ) , then ( ?? - ?? h ) 2 = ?? 2 ( 1 + ?? 2 ) or ?? 2 ( h 2 - ?? 2 ) - 2 ?? h ?? + ( ?? 2 - ?? 2 ) = 0, where ?? = tan ? ?? c ot ? ?? 1 + c ot ? ?? 2 = ?? ? 1 ?? 1 + 1 ?? 2 = ?? ? ?? 1 + ?? 2 ?? 1 ?? 2 = ?? ? ? 2 h ?? ( ?? 2 - ?? 2 ) = ?? ? ?? ( ?? 2 - ?? 2 ) = 2 ???? Q7: The chord of contact of tangents from a point ?? to a circle passes through ?? . If ?? ?? and ?? ?? are the lengths of the tangents from ?? and ?? to the circle, then ???? is equal to (A) ?? ?? + ?? ?? ?? (B) ?? ?? - ?? ?? ?? (C) v ?? ?? ?? + ?? ?? ?? (D) v ?? ?? ?? - ?? ?? ?? Ans: (C) Let ?? ( ?? 1 , ?? 1 ) and ?? ( ?? 2 , ?? 2 ) be two points and ?? 2 + ?? 2 = ?? 2 be the given circle then the chord of contact of tangents drawn from ?? to the given circle is xx 1 + yy 1 = a 2 It will pass through ?? ( ?? 2 , ?? 2 ) , if ? ?? 2 ?? 1 + ?? 2 ?? 1 = ?? 2 Now, ?? 1 = v ?? 1 2 + ?? 1 2 - ?? 2 , ? ?? 2 = v ?? 2 2 + ?? 2 2 - ?? 2 ? ? ???? = v ( ?? 2 - ?? 1 ) 2 + ( ?? 2 - ?? 1 ) 2 ? ? ???? = v( ?? 2 2 + ?? 2 2 ) + ( ?? 1 2 + ?? 1 2 ) - 2 ( ?? 1 ?? 2 + ?? 1 ?? 2 ) ? ? ???? = v( ?? 2 2 + ?? 2 2 ) + ( ?? 1 2 + ?? 1 2 ) - 2 ?? 2 ? [using (i)] ? ? ???? = v( ?? 1 2 + ?? 1 2 - ?? 2 ) + ( ?? 2 2 + ?? 2 2 - ?? 2 ) ? ? ???? = v ?? 1 2 + ?? 2 2 Page 5 JEE Solved Example on Straight Lines JEE Mains Q1: The opposite angular points of a square are ( ?? , ?? ) and ( ?? , - ?? ) . Then the co-ordinates of other two points are (A) D( ?? ?? , ?? ?? ) , ?? ( - ?? ?? , ?? ?? ) (B) ?? ( ?? ?? , ?? ?? ) , ?? ( ?? ?? , ?? ?? ) (C) ?? ( ?? ?? , ?? ?? ) , ?? ( - ?? ?? , ?? ?? ) (D) None of these Ans: (C) Obviously, slope of AC = 5 / 2. Let ?? be the slope of a line inclined at an angle of 45 ° to ???? , then tan ? 45 ° = ± ?? - 5 2 1 + ?? · 5 2 ? ?? = - 7 3 · 3 7 . Thus, let the slope of ???? or ???? be 3/7and that of ???? or ???? be - 7 3 . Then equation of ???? is 3x - 7y + 19 = 0. Also the equation of ???? is 7x + 3y - 4 = 0. On solving these equations, we get, B ( - 1 2 , 5 2 ). Now let the coordinates of the vertex ?? be ( h , ?? ) . Since the middle points of ???? and ???? are same, therefore 1 2 ( ?? - 1 2 ) = 1 2 ( 3 + 1 ) ? ?? = 9 2 , 1 2 ( ?? + 5 2 ) = 1 2 ( 4 - 1 ) ? ? ?? = 1 2 . Hence, ?? = ( 9 2 , 1 2 ) . Q2: If the lines ?? = ?? ?? + ?? and ?? ?? = ?? + ?? are equally inclined to the line ?? = ???? + ?? , then ?? = (A) ?? + ?? v ?? ?? (B) ?? - ?? v ?? ?? (C) ?? ± ?? v ?? ?? (D) ?? ± ?? v ?? ?? Ans: (D) ?? 1 = 3 , ?? 2 = 1 2 and ?? 3 = ?? Let the angle between first and third line is ?? , and between second and third is ?? 2 , then tan ? ?? 1 = 3 - ?? 1 + 3 ?? and tan ? ?? 2 = ?? - 1 2 1 + ?? 2 But ?? 1 = ?? 2 ? ? ? 3 - ?? 1 + 3 ?? = ?? - 1 2 1 + ?? 2 ? ? 7 ?? 2 - 2 ?? - 7 = 0 ? ?? = 1 ± 5 v 2 7 Q3: Let ?? ( - ?? , ?? ) , ?? ( ?? , ?? ) and ?? ( ?? , ?? v ?? ) be three points. Then the equation of the bisector of the angle ???? ?? is (A) v ?? ?? ?? + ?? = ?? (B) ?? + v ?? ?? = ?? (C) v ?? ?? + ?? = ?? (D) ?? + v ?? ?? ?? = ?? Ans: (C) Slope of QR = 3 v 3 - 0 3 - 0 = v 3 i.e., ?? = 60 ° Clearly, ? ?????? = 120 ° ???? is the angle bisector of the angle, so line ???? makes 120 ° with the positive direction of ?? -axis. Therefore, equation of the bisector of ? ?????? is ?? = tan ? 120 ° ?? or ?? = v 3 ?? i.e., v 3 ?? + ?? = 0. Q4: If for a variable line ?? ?? + ?? ?? = ?? , the condition ?? ?? ?? + ?? ?? ?? = ?? ?? ?? ? ( ?? is a constant) is satisfied, then locus of foot of perpendicular drawn from origin to the line is (A) ?? ?? + ?? ?? = ?? ?? / ?? (B) ?? ?? + ?? ?? = ?? ?? ?? (C) ?? ?? + ?? ?? = ?? ?? (D) ?? ?? - ?? ?? = ?? ?? Ans: (C) Equation of perpendicular drawn from origin to the line ?? ?? + ?? ?? = 1 is ?? - 0 = ?? ?? ( ?? - 0 ) [ ? m of given line = - ?? ?? , ? m of perpendicular = ?? ?? ] ? by - ax = 0 ? ?? ?? - ?? ?? = 0 Now, the locus of foot of perpendicular is the intersection point of line ?? ?? + ?? ?? = 1. and ?? ?? - ?? ?? = 0. To find locus, squaring and adding (i) and (ii) ( ?? ?? + ?? ?? ) 2 + ( ?? ?? - ?? ?? ) 2 = 1 ? ?? 2 ( 1 ?? 2 + 1 ?? 2 ) + ?? 2 ( 1 ?? 2 + 1 ?? 2 ) = 1 ? ?? 2 ( 1 ?? 2 ) + ?? 2 ( 1 ?? 2 ) = 1 , [ ? 1 ?? 2 + 1 ?? 2 = 1 ?? 2 ] ? ?? 2 + ?? 2 = ?? 2 . Q5: If the circle ?? ?? + ?? ?? - ?? ?? - ?? ?? + ?? = ?? does not touch or intersect the axes and the point ( ?? , ?? ) lies inside the circle, then (A) ?? < ?? < ?? (B) ?? < ?? < ???? (C) ?? < ?? < ???? (D) none of these Ans: (C) The centre of the given circle is ( 2 , 3 ) and the radius = v 4 + 9 - ?? i.e. v 13 - k. Since the given circle does not touch or intersect the coordinate axes and the point ( 2 , 2 ) lies inside the circle ? ?? -cooridnate of centre > radius i.e. 2 > v 13 - k, ?? -coordinate of centre > radius i.e. 3 > v 13 - k and 4 + 4 - 8 - 12 + ?? < 0 ? 4 > 13 - ?? , 9 > 13 - ?? and - 12 + ?? < 0 ? ?? > 9 , ?? > 4 and ?? < 12 ? 9 < ?? < 12 Q6: If ?? ?? , ?? ?? be the inclination of tangents drawn from the point ?? to the circle ?? ?? + ?? ?? = ?? ?? with ?? -axis, then the locus of ?? , it is given that ?? ???? ? ?? ?? + ?? ???? ? ?? ?? = ?? , is (A) ?? ( ?? ?? - ?? ?? ) = ?? ???? (B) ?? ( ?? ?? - ?? ?? ) = ?? ?? - ?? ?? (C) ?? ( ?? ?? - ?? ?? ) = ?? ???? (D) none of these Ans: (C) Any tangent is ?? = ???? ± ?? v 1 + ?? 2 If it passes through ?? ( h , ?? ) , then ( ?? - ?? h ) 2 = ?? 2 ( 1 + ?? 2 ) or ?? 2 ( h 2 - ?? 2 ) - 2 ?? h ?? + ( ?? 2 - ?? 2 ) = 0, where ?? = tan ? ?? c ot ? ?? 1 + c ot ? ?? 2 = ?? ? 1 ?? 1 + 1 ?? 2 = ?? ? ?? 1 + ?? 2 ?? 1 ?? 2 = ?? ? ? 2 h ?? ( ?? 2 - ?? 2 ) = ?? ? ?? ( ?? 2 - ?? 2 ) = 2 ???? Q7: The chord of contact of tangents from a point ?? to a circle passes through ?? . If ?? ?? and ?? ?? are the lengths of the tangents from ?? and ?? to the circle, then ???? is equal to (A) ?? ?? + ?? ?? ?? (B) ?? ?? - ?? ?? ?? (C) v ?? ?? ?? + ?? ?? ?? (D) v ?? ?? ?? - ?? ?? ?? Ans: (C) Let ?? ( ?? 1 , ?? 1 ) and ?? ( ?? 2 , ?? 2 ) be two points and ?? 2 + ?? 2 = ?? 2 be the given circle then the chord of contact of tangents drawn from ?? to the given circle is xx 1 + yy 1 = a 2 It will pass through ?? ( ?? 2 , ?? 2 ) , if ? ?? 2 ?? 1 + ?? 2 ?? 1 = ?? 2 Now, ?? 1 = v ?? 1 2 + ?? 1 2 - ?? 2 , ? ?? 2 = v ?? 2 2 + ?? 2 2 - ?? 2 ? ? ???? = v ( ?? 2 - ?? 1 ) 2 + ( ?? 2 - ?? 1 ) 2 ? ? ???? = v( ?? 2 2 + ?? 2 2 ) + ( ?? 1 2 + ?? 1 2 ) - 2 ( ?? 1 ?? 2 + ?? 1 ?? 2 ) ? ? ???? = v( ?? 2 2 + ?? 2 2 ) + ( ?? 1 2 + ?? 1 2 ) - 2 ?? 2 ? [using (i)] ? ? ???? = v( ?? 1 2 + ?? 1 2 - ?? 2 ) + ( ?? 2 2 + ?? 2 2 - ?? 2 ) ? ? ???? = v ?? 1 2 + ?? 2 2 Q8: The circles ?? ?? + ?? ?? + ?? ?? ?? ?? - ?? ?? = ?? and ?? ?? + ?? ?? + ?? ?? ?? ?? - ?? ?? = ?? cut each other orthogonally. If ?? ?? , ?? ?? are perpendicular from ( ?? , ?? ) and ( ?? , - ?? ) on a common tangent of these circle, then ?? ?? ?? ?? is equal to (A) ?? ?? ?? (B) ?? ?? (C) ?? ?? ?? (D) ?? ?? + ?? Ans: (B) Given ?? 1 ?? 2 + a 2 = 0 If I + my = 1 is common tangent of these circle, then - | ?? 1 - 1 v 1 1 + ?? 2 = ± v ?? 1 2 + ?? 2 = ( ?? ?? 1 + 1 ) 2 = ( 1 2 + ?? 2 ) ( ?? 1 2 + ?? 2 ) ? ? ?? 2 ?? 1 2 - 2 ?? ?? 1 + ?? 2 ( 1 2 + ?? 2 ) - 1 = 0 Similarly, ?? 2 ?? 2 2 - 21 ?? 2 + ?? 2 ( 1 2 + ?? 2 ) - 1 = 0 So that g 1 , g 2 are the roots of the equation ?? 2 ?? 2 - 2 ???? + ?? 2 ( ?? 2 + ?? 2 ) - 1 = 0 ? ? ?? 1 ?? 2 = ?? 2 ( ?? 2 + ?? 2 ) - 1 ?? 2 = - ?? 2 ? ? ?? 2 ( ?? 2 + ?? 2 ) = 1 - ?? 2 ?? 2 ? … Now, ?? 1 ?? 2 = | ???? - 1 | v 1 2 + ?? 2 · | - ???? - 1 | v ?? 2 + ?? 2 = | 1 - ?? 2 ?? 2 | 1 2 + ?? 2 = ?? 2 [ from (ii)] Q9: The equation of the line touching both the parabolas ?? ?? = ?? ?? and ?? ?? = - ???? ?? is (A) ?? ?? + ?? - ?? = ?? (B) ?? + ?? ?? + ?? = ?? (C) ?? - ?? ?? + ?? = ?? (D) ?? - ?? ?? - ?? = ?? Ans: (C) Any tangent to ?? 2 = 4 ?? is ?? = ???? + 1 ?? Solving with ?? 2 = - 32 ?? ?? 2 = - 32 ( ???? + 1 ?? ) ? ?? 2 + 32 ???? + 32 ?? = 0 But the line is tangent to 2 nd parabola. Discriminant = 0 ? ( 32 ) 2 ?? 2 = 4 · 32 ?? ? ? ? ?? 3 = 4 . 32 ( 32 ) 2 = 1 8 ? ? ?? = 1 2Read More
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FAQs on Solved Example: Straight Lines - Mathematics (Maths) for JEE Main & Advanced
| 1. How to find the equation of a straight line passing through two given points? |  |
Ans. To find the equation of a straight line passing through two given points, first calculate the slope using the formula (y2 - y1)/(x2 - x1). Then, substitute one of the points and the slope into the point-slope form equation y - y1 = m(x - x1).
| 2. What is the condition for two lines to be parallel? | |
Ans. Two lines are parallel if they have the same slope but different y-intercepts. This means that the slopes of the two lines are equal, but the y-intercepts are not equal.
| 3. How to determine if two lines are perpendicular to each other? | |
Ans. Two lines are perpendicular if the product of their slopes is -1. In other words, if the slopes of the two lines are negative reciprocals of each other, then they are perpendicular.
| 4. How to find the distance between a point and a line? | |
Ans. To find the distance between a point (x1, y1) and a line Ax + By + C = 0, use the formula |Ax1 + By1 + C| / √(A^2 + B^2).
| 5. What is the significance of the slope of a straight line? | |
Ans. The slope of a straight line represents the rate at which the line is rising or falling. It also indicates the direction of the line - a positive slope means the line is increasing, while a negative slope means the line is decreasing.
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Solved Example: Straight Lines Notes
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The content of Solved Example: Straight Lines is prepared as per the latest JEE syllabus.
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