1. Chapter 10 Class 11 Straight Lines
2. Serial order wise

Transcript

Example 23 Show that the area of the triangle formed by the lines y = m1 x + c1 , y = m2x + c2 and x = 0 is (๐_2 โ ๐_1 )^2/2|๐_1 โ ๐_2 | . There are three lines given in the graph x = 0 This line will lie be y-axis y = m1 x + c1 Putting x = 0 y = 0 + c1 = c1 Hence point is P(0, c1) y = m2 x + c2 Putting x = 0 y = 0 + c2 = c2 Hence point is Q(0, c2) Assuming both lines meet at R We need to find area ฮ PQR To find area ฮ PQR, we find coordinates of vertices Here P(0, c1), Q(0,c2) We find coordinates of point R Vertex R is the point of intersection of lines y = m1 x + c1 โฆ(1) y = m2 x + c2 โฆ(2) Subtracting (1) from (2) y โ y = m1 x + c1 โ (m2x + c2) 0 = m1 x + c1 โ m2x โ c2 0 = x(m1 โ m2) + c1 โ c2 โx(m1 โ m2) = c1 โ c2 x(m1 โ m2) = โ(c1 โ c2) x(m1 โ m2) = c2 โ c1 x = (๐_2 โ ๐_1)/(๐_1 โ ใ ๐ใ_2 ) Putting value of x in (1) y = m1x + c1 y = m1 ((๐_2 โ ๐_1)/(๐_1 โ ๐_2 )) + c1 y = (๐_1 (๐_2 โ ๐_1 ) + ใ ๐ใ_1 (๐_1 โ ๐_2))/(๐_1 โ ๐_2 ) y = (๐_1 ๐_2 โ ๐_1 ๐_1 + ใ ๐ใ_1 ๐_1 โ ใ ๐ใ_1 ๐_2)/(๐_1 โ ๐_2 ) y = (๐_1 ๐_(2 ) โ ใ ๐ใ_1 ๐_2)/(๐_1 โ ๐_2 ) โด Vertex R is ((๐_2 โ ๐_1)/(๐_1 โ ๐_2 ) ", " (๐_1 ๐_(2 ) โ ๐_2 ใ ๐ใ_1)/(๐_1 โ ๐_2 )) The vertices of โ PRQ is P(0, c1), R((๐_2 โ ๐_1)/(๐_1 โ ๐_2 ) ", " (๐_1 ๐_(2 ) โ ๐_2 ใ ๐ใ_1)/(๐_1 โ ๐_2 )) & Q(0, c2) We know that Area of triangle whose vertices are (x1, y1) (x2, y2) (x3, y3) is 1/2 |"x1" ("y2 โ y3" ) + ๐ฅ2 (๐ฆ3 โ ๐ฆ1) + ๐ฅ3(๐ฆ1 โ ๐ฆ2)| For โ PRQ, (x1, y1) = P(0, c1) (x2,y2) = R ((๐_2 โ ๐_1)/(๐_1 โ ๐_2 ) ", " (๐_1 ๐_(2 ) โ ๐_2 ใ ๐ใ_1)/(๐_1 โ ๐_2 )) (x3,y3) = Q (0, c2) Area โ PRQ = โ 8(1/2 " " |0((๐_1 ๐_(2 ) โ ๐_2 ใ ๐ใ_1)/(๐_1 โ ๐_2 ) โ๐2) + (๐_2 โ ๐_1)/(๐_1 โ ๐_2 ) (๐2 โ๐1) + 0(๐1 โ (๐_1 ๐_(2 ) โ ๐_2 ใ ๐ใ_1)/(๐_1 โ ๐_2 ))| ) = โ 8(1/2 " " |โ(0 + (๐2 โ ๐1)2/(๐1 โ ๐2) + 0)| ) = โ 8(1/2 " " |(ใ(๐ใ_2 โ ๐_1)2)/(๐_1 โ ๐_2 )| ) = 1/2 (๐2 โ ๐1)2/|๐1 โ ๐2| โด Area required = 1/2 (๐2 โ ๐1)2/|๐1 โ ๐2| Hence proved