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Example 23 - Show that area of triangle formed by y = m1x + c1 - Other Type of questions - Area of Triangle formed

  1. Chapter 10 Class 11 Straight Lines
  2. Serial order wise
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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 โˆ’ ๐‘_1 )^2/2|๐‘š_1 โˆ’ ๐‘š_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 โˆ’ ๐‘_1)/(๐‘š_1 โˆ’ ใ€– ๐‘šใ€—_2 ) Putting value of x in (1) y = m1x + c1 y = m1 ((๐‘_2 โˆ’ ๐‘_1)/(๐‘š_1 โˆ’ ๐‘š_2 )) + c1 y = (๐‘š_1 (๐‘_2 โˆ’ ๐‘_1 ) + ใ€– ๐‘ใ€—_1 (๐‘š_1 โˆ’ ๐‘š_2))/(๐‘š_1 โˆ’ ๐‘š_2 ) y = (๐‘š_1 ๐‘_2 โˆ’ ๐‘š_1 ๐‘_1 + ใ€– ๐‘ใ€—_1 ๐‘š_1 โˆ’ ใ€– ๐‘ใ€—_1 ๐‘š_2)/(๐‘š_1 โˆ’ ๐‘š_2 ) y = (๐‘š_1 ๐‘_(2 ) โˆ’ ใ€– ๐‘ใ€—_1 ๐‘š_2)/(๐‘š_1 โˆ’ ๐‘š_2 ) โˆด Vertex R is ((๐‘_2 โˆ’ ๐‘_1)/(๐‘š_1 โˆ’ ๐‘š_2 ) ", " (๐‘š_1 ๐‘_(2 ) โˆ’ ๐‘š_2 ใ€– ๐‘ใ€—_1)/(๐‘š_1 โˆ’ ๐‘š_2 )) The vertices of โˆ† PRQ is P(0, c1), R((๐‘_2 โˆ’ ๐‘_1)/(๐‘š_1 โˆ’ ๐‘š_2 ) ", " (๐‘š_1 ๐‘_(2 ) โˆ’ ๐‘š_2 ใ€– ๐‘ใ€—_1)/(๐‘š_1 โˆ’ ๐‘š_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 โˆ’ ๐‘ฆ1) + ๐‘ฅ3(๐‘ฆ1 โˆ’ ๐‘ฆ2)| For โˆ† PRQ, (x1, y1) = P(0, c1) (x2,y2) = R ((๐‘_2 โˆ’ ๐‘_1)/(๐‘š_1 โˆ’ ๐‘š_2 ) ", " (๐‘š_1 ๐‘_(2 ) โˆ’ ๐‘š_2 ใ€– ๐‘ใ€—_1)/(๐‘š_1 โˆ’ ๐‘š_2 )) (x3,y3) = Q (0, c2) Area โˆ† PRQ = โ– 8(1/2 " " |0((๐‘š_1 ๐‘_(2 ) โˆ’ ๐‘š_2 ใ€– ๐‘ใ€—_1)/(๐‘š_1 โˆ’ ๐‘š_2 ) โˆ’๐‘2) + (๐‘_2 โˆ’ ๐‘_1)/(๐‘š_1 โˆ’ ๐‘š_2 ) (๐‘2 โˆ’๐‘1) + 0(๐‘1 โˆ’ (๐‘š_1 ๐‘_(2 ) โˆ’ ๐‘š_2 ใ€– ๐‘ใ€—_1)/(๐‘š_1 โˆ’ ๐‘š_2 ))| ) = โ– 8(1/2 " " |โ–ˆ(0 + (๐‘2 โˆ’ ๐‘1)2/(๐‘š1 โˆ’ ๐‘š2) + 0)| ) = โ– 8(1/2 " " |(ใ€–(๐‘ใ€—_2 โˆ’ ๐‘_1)2)/(๐‘š_1 โˆ’ ๐‘š_2 )| ) = 1/2 (๐‘2 โˆ’ ๐‘1)2/|๐‘š1 โˆ’ ๐‘š2| โˆด Area required = 1/2 (๐‘2 โˆ’ ๐‘1)2/|๐‘š1 โˆ’ ๐‘š2| Hence proved

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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