Short Answer TypeIn Figure, the vertices of Δ ABC are A(4, 6), B(1, 5) and C(7, 2). A line segment DE is drawn to intersect the sides AB and AC at D and E respectively such that 
Calculate the area of Δ ADE and compare it with an area of Δ ABC.
The points A (4, 7), B (p, 3) and C (7, 3) are the vertices a right triangle, right-angled at B find the value of p.
Find the relation between x and y if the points A (x, y), B (-5, 7) and C (-4, 5) are collinear.
If the coordinates of points A and B are (-2, -2) and (2, -4) respectively, find the coordinates of P such that AP = 3/7 AB, where P lies on the line segment AB.
Long Answer TypeFind the values of k so that the area of the triangle with vertices (1,-1), (-4, 2k) and (-k, 5) is 24 sq. units
Short Answer TypeIf the distance between the points (4, k) and (1, 0) is 5, then what can be the possible values of k?
The area of a triangle is 5 sq units. Two of its vertices are (2, 1) and (3, –2). If the third vertex is (7/2, y), find the value of y.
Long Answer TypeIf a≠b≠0, prove that the points (a, a2), (b, b2) (0, 0) will not be collinear.
Let A(a, a2), B(b, b2) and C(0, 0) be the coordinates of the given points.
We know that the area of a triangle having vertices (x1, y1), (x2, y2) and (x3, y3) is ∣∣12[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]∣∣ square units.
So,
Area of ∆ABC\
Since the area of the triangle formed by the points (a, a2), (b, b2) and (0, 0) is not zero, so the given points are not collinear.
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