Check sibling questions

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


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 (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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.