Find the area of a triangle whose vertices are A(3,2), B(11,6), and C(7,14).
40 sq. units
Area of a triangle with vertices \((x_1,y_1), (x_2,y_2), (x_3,y_3)\) is \(\dfrac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|\).
\(= \dfrac{1}{2}|3(6-14)+11(14-2)+7(2-6)| = \dfrac{1}{2}|-24+132-28| = \dfrac{1}{2} \times 80 = 40\).
Hence, the correct answer is 40 sq. units (D).
Important Questions from Co-ordinate Geometry
The graphs of the linear equations 4x - 2y = 10 and 4x + ky = 2 intersect at a point (a, 4). The value of k is equal to:
In which ratio the point (-3, p) divides the line segment joining the points (-5, -4) and (-2, 3)?
The area (in sq. units) of the triangle formed by the graphs of 8x + 3y = 24, 2x + 8 = y and the x-axis is:
In which quadrant both abscissa and ordinate are negative?
Find the slope of the line joining the points (3, -4) and (5, 2).
