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.
Given ,
The vertices of a right triangle, such as,
A (4, 7), B (p, 3) and C (7, 3) 
In right ΔABC, using Pythagoras theorem
(AB)2 + (BC)2 = (AC)2
⇒ [(3 - 7)2 + (p - 4)2] + [(3-3)2 +(7 - p)2] = [(3-7)2 + (7 - 4)2]
⇒ (p - 4)2 + (7 - p)2 = 9
⇒ p2 + 16 -8p + 49 + p2 - 14p = 9
⇒ 2p2 - 22p + 56 = 0
⇒ p2 - 11p + 28 = 0
⇒ p2 - 4p -7p + 28 =0
⇒ p(p - 4) -7(p - 4)=0
⇒ (p - 7 )(p - 4) = 0
⇒ p - 7 = 0 or p - 4 = 0
⇒ p = 7 or p = 4
Hence, the value of p is 4 or 7
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 TypeShort Answer Type
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