Advertisements
Advertisements
प्रश्न
Find the area of the triangle whose vertices are (–8, 4), (–6, 6) and (–3, 9).
उत्तर
Given that, the vertices of triangles are (–8, 4), (–6, 6) and (–3, 9).
Let (x1, y1) `→` (−8, 4)
(x2, y2) `→` (−6, 6)
And (x3, y3) `→` (−3, 9)
We know that, the area of triangle with vertices
(x1, y1), (x2, y2) and (x3, y3)
Δ = `1/2[x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)]`
= `1/2[-8(6 - 9) - 6(9 - 4) + (-3)(4 - 6)]`
= `1/2[-8(-3) - 6(5) - 3(-2)]`
= `1/2(24 - 30 + 6)`
= `1/2(30 - 30)`
= `1/2(0)`
= 0
Hence, the required area of triangle is 0.
APPEARS IN
संबंधित प्रश्न
Find the relation between x and y if, the points A(x, y), B(-5, 7) and C(-4, 5) are collinear.
Find the values of k so that the area of the triangle with vertices (k + 1, 1), (4, -3) and (7, -k) is 6 sq. units.
If `a≠ b ≠ c`, prove that the points (a, a2), (b, b2), (c, c2) can never be collinear.
prove that the points A (7, 10), B(-2, 5) and C(3, -4) are the vertices of an isosceles right triangle.
Show that the points (-3, -3),(3,3) and C (-3 `sqrt(3) , 3 sqrt(3))` are the vertices of an equilateral triangle.
The area of a triangle with vertices (–3, 0), (3, 0) and (0, k) is 9 sq.units. The value of k will be ______.
The area of a triangle with base 4 cm and height 6 cm is 24 cm2.
Area of a right-angled triangle is 30 cm2. If its smallest side is 5 cm, then its hypotenuse is ______.
In the given figure, area of ΔPQR is 20 cm2 and area of ΔPQS is 44 cm2. Find the length RS, if PQ is perpendicular to QS and QR is 5 cm.
If (a, b), (c, d) and (e, f) are the vertices of ΔABC and Δ denotes the area of ΔABC, then `|(a, c, e),(b, d, f),(1, 1, 1)|^2` is equal to ______.
