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

Example 23 - Chapter 10 Class 11 Straight Lines - Part 2
Example 23 - Chapter 10 Class 11 Straight Lines - Part 3
Example 23 - Chapter 10 Class 11 Straight Lines - Part 4
Example 23 - Chapter 10 Class 11 Straight Lines - Part 5

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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 (𝑐_2 βˆ’ 𝑐_1 )^2/2|π‘š_1 βˆ’ π‘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 Ξ” PQR To find area Ξ” 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 …(1) y = m2 x + c2 …(2) Subtracting (1) from (2) y – y = m1 x + c1 – (m2x + c2) 0 = m1 x + c1 – m2x – c2 0 = x(m1 – m2) + c1 – c2 –x(m1 – m2) = c1 – c2 x(m1 – m2) = –(c1 – c2) x(m1 – m2) = c2 – c1 x = (𝑐_2 βˆ’ 𝑐_1)/(π‘š_1 βˆ’ γ€– π‘šγ€—_2 ) Putting value of x in (1) y = m1x + c1 y = m1 ((𝑐_2 βˆ’ 𝑐_1)/(π‘š_1 βˆ’ π‘š_2 )) + c1 y = (π‘š_1 (𝑐_2 βˆ’ 𝑐_1 ) + γ€– 𝑐〗_1 (π‘š_1 βˆ’ π‘š_2))/(π‘š_1 βˆ’ π‘š_2 ) y = (π‘š_1 𝑐_2 βˆ’ π‘š_1 𝑐_1 + γ€– 𝑐〗_1 π‘š_1 βˆ’ γ€– 𝑐〗_1 π‘š_2)/(π‘š_1 βˆ’ π‘š_2 ) y = (π‘š_1 𝑐_(2 ) βˆ’ γ€– 𝑐〗_1 π‘š_2)/(π‘š_1 βˆ’ π‘š_2 ) ∴ Vertex R is ((𝑐_2 βˆ’ 𝑐_1)/(π‘š_1 βˆ’ π‘š_2 ) ", " (π‘š_1 𝑐_(2 ) βˆ’ π‘š_2 γ€– 𝑐〗_1)/(π‘š_1 βˆ’ π‘š_2 )) The vertices of βˆ† PRQ is P(0, c1), R((𝑐_2 βˆ’ 𝑐_1)/(π‘š_1 βˆ’ π‘š_2 ) ", " (π‘š_1 𝑐_(2 ) βˆ’ π‘š_2 γ€– 𝑐〗_1)/(π‘š_1 βˆ’ π‘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 – y3" ) + π‘₯2 (𝑦3 βˆ’ 𝑦1) + π‘₯3(𝑦1 βˆ’ 𝑦2)| For βˆ† PRQ, (x1, y1) = P(0, c1) (x2,y2) = R ((𝑐_2 βˆ’ 𝑐_1)/(π‘š_1 βˆ’ π‘š_2 ) ", " (π‘š_1 𝑐_(2 ) βˆ’ π‘š_2 γ€– 𝑐〗_1)/(π‘š_1 βˆ’ π‘š_2 )) (x3,y3) = Q (0, c2) Area βˆ† PRQ = β– 8(1/2 " " |0((π‘š_1 𝑐_(2 ) βˆ’ π‘š_2 γ€– 𝑐〗_1)/(π‘š_1 βˆ’ π‘š_2 ) βˆ’π‘2) + (𝑐_2 βˆ’ 𝑐_1)/(π‘š_1 βˆ’ π‘š_2 ) (𝑐2 βˆ’π‘1) + 0(𝑐1 βˆ’ (π‘š_1 𝑐_(2 ) βˆ’ π‘š_2 γ€– 𝑐〗_1)/(π‘š_1 βˆ’ π‘š_2 ))| ) = β– 8(1/2 " " |β–ˆ(0 + (𝑐2 βˆ’ 𝑐1)2/(π‘š1 βˆ’ π‘š2) + 0)| ) = β– 8(1/2 " " |(γ€–(𝑐〗_2 βˆ’ 𝑐_1)2)/(π‘š_1 βˆ’ π‘š_2 )| ) = 1/2 (𝑐2 βˆ’ 𝑐1)2/|π‘š1 βˆ’ π‘š2| ∴ Area required = 1/2 (𝑐2 βˆ’ 𝑐1)2/|π‘š1 βˆ’ π‘š2| Hence proved

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.