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Examples

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Example 11 Deleted for CBSE Board 2023 Exams

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Example 14 Important Deleted for CBSE Board 2023 Exams

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Example 23 You are here

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Last updated at Dec. 8, 2016 by Teachoo

Example 23 Show that the area of the triangle formed by the lines y = m1 x + c1 , y = m2x + c2 and x = 0 is (π_2 β π_1 )^2/2|π_1 β π_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 β¦(1) y = m2 x + c2 β¦(2) Subtracting (1) from (2) y β y = m1 x + c1 β (m2x + c2) 0 = m1 x + c1 β m2x β c2 0 = x(m1 β m2) + c1 β c2 βx(m1 β m2) = c1 β c2 x(m1 β m2) = β(c1 β c2) x(m1 β m2) = c2 β c1 x = (π_2 β π_1)/(π_1 β γ πγ_2 ) Putting value of x in (1) y = m1x + c1 y = m1 ((π_2 β π_1)/(π_1 β π_2 )) + c1 y = (π_1 (π_2 β π_1 ) + γ πγ_1 (π_1 β π_2))/(π_1 β π_2 ) y = (π_1 π_2 β π_1 π_1 + γ πγ_1 π_1 β γ πγ_1 π_2)/(π_1 β π_2 ) y = (π_1 π_(2 ) β γ πγ_1 π_2)/(π_1 β π_2 ) β΄ Vertex R is ((π_2 β π_1)/(π_1 β π_2 ) ", " (π_1 π_(2 ) β π_2 γ πγ_1)/(π_1 β π_2 )) The vertices of β PRQ is P(0, c1), R((π_2 β π_1)/(π_1 β π_2 ) ", " (π_1 π_(2 ) β π_2 γ πγ_1)/(π_1 β π_2 )) & Q(0, c2) We know that Area of triangle whose vertices are (x1, y1) (x2, y2) (x3, y3) is 1/2 |"x1" ("y2 β y3" ) + π₯2 (π¦3 β π¦1) + π₯3(π¦1 β π¦2)| For β PRQ, (x1, y1) = P(0, c1) (x2,y2) = R ((π_2 β π_1)/(π_1 β π_2 ) ", " (π_1 π_(2 ) β π_2 γ πγ_1)/(π_1 β π_2 )) (x3,y3) = Q (0, c2) Area β PRQ = β 8(1/2 " " |0((π_1 π_(2 ) β π_2 γ πγ_1)/(π_1 β π_2 ) βπ2) + (π_2 β π_1)/(π_1 β π_2 ) (π2 βπ1) + 0(π1 β (π_1 π_(2 ) β π_2 γ πγ_1)/(π_1 β π_2 ))| ) = β 8(1/2 " " |β(0 + (π2 β π1)2/(π1 β π2) + 0)| ) = β 8(1/2 " " |(γ(πγ_2 β π_1)2)/(π_1 β π_2 )| ) = 1/2 (π2 β π1)2/|π1 β π2| β΄ Area required = 1/2 (π2 β π1)2/|π1 β π2| Hence proved