d. The area of A ABC is 5 sq units. Two of its vertices are A(2, 1) and B (3, -2).
Third vertex C lies on the line given by y-x + 3 = 0. Find the coordinates of C.
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Step-by-step explanation:
Let A(2, 1) and B(3, -2) be the vertices of Δ And C(x,y) be the third vertex Area of ΔABC But it is given that area of ΔABC=5 But it is given that third vertices lies on y = x +3 Hence subsisting value of y in (i) 3x + x +3 =17 or 3x + x +3 = -3 Hence coordinates of c will be (7/2,13/2)
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