Find the area of the triangle whose vertices are:

(i) (2, 3), (-1, 0), (2, -4)

(ii) (-5, -1), (3, -5), (5, 2)

Asked by Pragya Singh | 1 year ago |  64

##### Solution :-

Area of a triangle formula = $$\dfrac{1}{2}$$× [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]

(i) Here,

x1 = 2, x2 = -1, x3 = 2, y1 = 3, y2 = 0 and y3 = -4

Substitute all the values in the above formula, we get

Area of triangle = $$\dfrac{1}{2}$$[2 {0- (-4)} + (-1) {(-4) – (3)} + 2 (3 – 0)]

=$$\dfrac{1}{2}$$ {8 + 7 + 6}

=$$\dfrac{21}{2}$$

So, area of triangle is $$\dfrac{21}{2}$$square units.

(ii) Here,

x1 = -5, x2 = 3, x3 = 5, y1 = -1, y2 = -5 and y3 = 2

Area of the triangle = $$\dfrac{1}{2}$$[-5 { (-5)- (2)} + 3(2-(-1)) + 5{-1 – (-5)}]

= $$\dfrac{1}{2}$${35 + 9 + 20} = 32

Therefore, the area of the triangle is 32 square units.

Answered by Abhisek | 1 year ago

### Related Questions

#### In the determine whether the given quadratic equations have real roots and if so, And the roots 3x2 – 2x + 2 = 0

In the determine whether the given quadratic equations have real roots and if so, And the roots 3x2 – 2x + 2 = 0

#### Find the point on x-axis which is equidistant from the points (-2, 5) and (2, -3).

Find the point on x-axis which is equidistant from the points (-2, 5) and (2, -3).

#### Show that the points A (5, 6), B (1, 5), C (2, 1) and D (6, 2) are the vertices of a square.

(i) Show that the points A (5, 6), B (1, 5), C (2, 1) and D (6, 2) are the vertices of a square.

(ii) Prove that the points A (2, 3), B (-2, 2), C (-1, -2) and D (3, -1) are the vertices of a square ABCD.

(iii) Name the type of triangle PQR formed by the point $$P(\sqrt{2} , \sqrt{2}), Q(- \sqrt{2}, – \sqrt{2)} and\; R (-\sqrt{6} , \sqrt{6} )$$