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# Find the area of the triangle formed by the lines $y – x = \theta, x + y = \theta$ and $x – k = \theta$

$\begin {array} {1 1} (A)\;k\: sq\: units & \quad (B)\;2k\: sq\: units \\ (C)\;k^2\: sq\: units & \quad (D)\;2k^2\: sq\: units \end {array}$

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• Area of a triangle whose vertices are $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$ is $\large\frac{1}{2}$$| x_1 (y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)| Equations of the given lines are : \qquad y-x=0---------(1) \qquad x+y=0----------(2) \qquad x-k=0----------(3) Let us now find the points of intersection of the above lines Solving equation (1) and (2) we get, \qquad x+y=0 \qquad y-x=0 We get x=0, y=0 Hence the point of intersection of line (1) and (2) is (0,0) Next let us solve equation (2) and (3) \qquad x+y=0 \qquad x = k \: \: \Rightarrow y = -k Hence the point of intersection of equation (2) and (3) is (k, -k) Next let us solve equation (3) and (1) \qquad x-k=0 \qquad y-x=0 \qquad y-k=0 \Rightarrow y=k also x = k \therefore The point of intersection is (k, k) Hence the vertices of the triangle formed by the three lines are (0,0) , (k, -k) and (k, k) Now substituting the values in area = \large\frac{1}{2}$$ | x_1 (y_2-y_1)+x_2(y_3-y_1)+x_3(y_1-y_2)|$ sq. units
area = $\large\frac{1}{2} | 0(-k \: -k)+k(k-0)+k(0+k)|$ square units
$= \large\frac{1}{2}|k^2+k^2|$ sq units

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