Tiêu đề XI - maths - chapter 10 - the straight line-ii.pdf ✅

1 Nishith Multimedia India (Pvt.) Ltd. 1 IX-Mathematics e-techno Text Book FAMILY OF LINES THROUGH THE INTERSECTION OF TWO GIVEN LINES THEOREM: Prove that the equation of the family of lines passing through the intersection of the lines a x b y c 1 1 1    0 and a x b y c 2 2 2    0 is a x b y c a x b y c 1 1 1 2 2 2           0 , where  is a parameter. PROOF : The equations of the lines are a x b y c 1 1 1    0 ....(i) and a x b y c 2 2 2    0 .....(ii) Let  ,  be the point of intersection of the lines (i) and (ii). then, a b c 1 1 1      0 .....(iii) and , a b c 2 2 2      0 ......(iv) Now, consider the equation a x b y c a x b y c 1 1 1 2 2 2           0 ......(v) Clearly, this is a first degree equation in x and y. So it represents a straight line. We have, a b c a b c 1 1 1 2 2 2              = 0 0 0    [ Using (iii) and (iv)] So,  ,  lies on the line given in (v). Hence, (v) represents family lines through the point of intersection of (i) and (ii). Thus, the family of straight lines through the intersection of lines L a x b y c 1 1 1 1     0 and L a x b y c 2 2 2 2     0 is a x b y c a x b y c 1 1 1 2 2 2           0 i.e., L L 1 2    0 . REMARK: The equation L L 1 2    0 represents a line passing through the intersection of the lines L1 =0 and L2 =0 which is a fixed point. Hence, L L 1 2    0 represents a family of straight lines, for different values of , which pass through a fixed - point. Example: Find the equation of the straight line which passes through the point (2,-3) and the point of intersection of the lines x + y + 4 = 0 and x - y - 8 = 0. Solution Any line through the intersection of the lines x + y + 4 = 0 and 3 x - y - 8 = 0. has the equation x y x y        4 3 8 0    ....(i)


4 Nishith Multimedia India (Pvt.) Ltd. 4 IX-Mathematics e-techno Text Book       1 2 2 1 2 2 1 2 2 1 1 2 1 2 1 2 1 2 a b a b tan a b a b a a b b a a b b          2 2 u tan u v v       where u a b a b   1 2 2 1 and v a a b b   1 2 1 2 2 2 u tan u v v      Now, two case arise : CASEI : When v a a b b    1 2 1 2 0: in this case, we have v  0    v 0 Also, 2 2 2 u v u u    2 2     u v v u    tan 1  Equation (iii) gives the bisector of the acute angle between the given lines. CASEII: When v a a b b    1 2 1 2 0 We know that 2 2 u v u v    2 2    u v v u    v v    2 2 1 u u v v        tan 1  Equation (iii) is the bisector of the obtuse angle between the lines (i) and (ii). Thus, if 1 2 c ,c are positive and a a b b 1 2 1 2   0, then 1 1 1 2 2 2 2 2 2 2 1 1 2 2 a x b y c a x b y c a b a b        is the bisector of acute angle between the lines and consequently. 1 1 1 2 2 2 2 2 2 2 1 1 2 2 a x b y c a x b y c a b a b         is the bisector of the obtuse angle between the lines. Also, if 1 2 c ,c are positive and a a b b 1 2 1 2   0, then 1 1 1 2 2 2 2 2 2 2 1 1 2 2 a x b y c a x b y c a b a b        is the bisector of the obtuse angle between the lines and consequently 1 1 1 2 2 2 2 2 2 2 1 1 2 2 a x b y c a x b y c a b a b         is the bisector of acute between them. From the above discussion, we obtain the following algorithm to find the bisectors of acute and obtuse angles between the lines a x b y c 1 1 1    0 and a x b y c 2 2 2    0.