Advertisements
Advertisements
Question
Find the area of the triangle whose vertices are: (4, 5), (0, 7), (–1, 1)
Solution
Here, A(x1, y1) ≡ A(4, 5), B(x2, y2) ≡ B(0, 7), C(x3, y3) ≡ C(–1, 1)
Area of a triangle = `1/2|(x_1, y_1, 1),(x_2, y_2, 1),(x_3, y_3, 1)|`
∴ A(Δ ABC) = `1/2|(4, 5, 1),(0, 7, 1),(-1, 1, 1)|`
= `1/2[4(7 - 1) - 5(0 + 1) + 1(0 + 7)]`
∴ A(Δ ABC) = `1/2(24 - 5 + 7)`
= 13 sq.units.
APPEARS IN
RELATED QUESTIONS
Find the area of the triangle whose vertices are: (0, 5), (0, – 5), (5, 0)
Find the value of k, if the area of the triangle with vertices at A(k, 3), B(–5, 7), C(–1, 4) is 4 square units.
Find the area of the quadrilateral whose vertices are A(–3, 1), B(–2, –2), C(4, 1), D(2, 3).
Find the area of triangles whose vertices are A(−1, 2), B(2, 4), C(0, 0).
Find the area of triangles whose vertices are L(1, 1), M(−2, 2), N(5, 4)
Find the value of k, if area of ΔPQR is 4 square units and vertices are P(k, 0), Q(4, 0), R(0, 2).
The maximum area of a right angled triangle with hypotenuse h is:
Find the values of 'K' if area of triangle is 4 square units and vertices are (k, 0), (4, 0), (0, 2).
If the area of triangle is 35 square units with vertices (2, – 6), (5, 4) and (k, 4) then k is
If Δ = `|(1, 2, 3),(2, -1, 0),(3, 4, 5)|`, then `|(1, 6, 3),(4, -6, 0),(3, 12, 5)|` is
Find the area of triangle whose vertices are A ( -1,2) ,B (2,4) ,C (0,0)
Find the area of triangle whose vertices are A(-1,2), B(2,4), C(0,0)
Find the area of triangle whose vertices are A(−1, 2), B(2, 4), C(0, 0).
Find the area of triangle whose vertices are A(-1, 2), B(2, 4), C(0, 0).
Find the area of triangle whose vertice are A(-1, 2), B(2, 4), C(0, 0)
Find the area of triangle whose vertices are A(−1, 2), B(2, 4), C(0, 0)
