Area of Triangle

Area of Triangle

Let (x1, y1), (x2, y2) and (x3, y3) respectively be the coordinates of the vertices A, B, C of a triangle ABC. Then the area of triangle ABC, is

...(1)

....(2)

While using formula (1) or (2). order of the points (xl, yL), (x2, y2) and (x3, y3) has not been taken into account. If we plot the points A(x1, y1), B (x2, y2) and C(x3, y3), then the area of the triangle as obtained by using formula (1) or (2) will be positive or negative as the points A, B, C are in anti-clockwise or clockwise directions.

while finding the area of triangle ABC, we use the formula :

Area of ΔABC    = Modulus of

Notes

(i) If three points (x1, y1), (x2, y2) and (x3, y3) are collinear, then

(ii) Equation of straight line passing through (x1, y1) and (x2, y2) is given by

(iii) In case of polygon with (x1, y1), (x2, y2) ............ (xn, yn) the area is given by

1/2 | (x1y 2 – y1x 2 ) + (x 2 y3 – y2x3 ) + ......... + (xn–1yn – yn–1xn ) + (xn y1 – yn x1) |

Ex.3 The vertices of quadrilateral in order are (–1, 4), (5, 6), (2, 9) and (x, x2). The area of the quadrilateral 15/2 sqft. units, then find the point (x, x2)

3x2 – 5x + 7 = ± 15

3x2 – 5x – 8 = 0,

3x2 – 5x + 22 = 0   ⇒  x = 8/3 , x = –1

Hence point is (8/3,64/9)  or (–1, 1). But (–1, 1) will not form a quadrilateral as per given order of the points. Hence the required point is  (8/3,64/9)

(1) Locus : When a point moves in a plane under certain geometrical conditions, the point traces out a path. This path of a moving point is called its locus.

Note : All those points which satisfy the given geometrical condition will definitely lie on the locus. But converse is not true always.

(2) Equation of Locus : The equation to a locus is the relation which exists between the coordinates of any point on the path, and which holds for no other point except those lying on the path.

(3) Procedure for finding the equation of the locus of a point

(i) If we are finding the equation of the locus of a point P, assign coordinates (h, k) to P.

(ii) Express the given conditions as equations in terms of the known quantities to facilitate calculations. We sometimes include some unknown quantities known as parameters.

(iii) Eliminate the parameters. so that the eliminate contains only h, k and known quantities.

(iv) Replace h by x, and k by y, in the eliminate. The resulting would be the equation of the locus of P. (v) If x and y coordinates of the moving point are obtained in terms of a third variable t (called the parameter), eliminate t to obtain the relation in x and y and simplify this relation. This will give the required equation of locus.

Ex. Find the focus of the middle points of the segment of a line passing through the point of intersection of the lines  ax + by + c = 0 and lx + my + n = 0 and intercepted between the axes.

Sol. Any line (say L = 0) passing through the point of intersection of ax + by + c = 0 and lx + my + n = 0  is (ax + by + c) + λ (lx + my + n) = 0, where l is any real number.

Point of intersection of L = 0 with axes are

Let the mid point be (h, k).

Eliminating λ    The required locus is : 2(am – lb) = (lc – an) x + (nb – mc)y..

(4) Inclination of a line : Its a measure of the smallest non-negative angle which the line makes with +ve direction of the x-axis [angle being measured in anti-clockwise direction]. 0 ≤ α < π

(5) Slope of the line : If the inclination of line is θ and  then its slope is defined as tan θ and denoted by 'm'

(i) If θ = 0, then m = 0 i.e. line parallel to x-axis.

(ii) If θ = 90º, then m does not exist i.e. line parallel to y-axis

(iii) Slope of line joining two points A(x1, y1) & B(x2, y2) is

(iv) If a line equally inclined with co-ordinate axes then slope is ± 1.

(6) Intercepts : The point where a line cuts the x-a xis (or y-a xis ) is called its x-intercep t (or y-intercept).

(i) Intercepts may be +ve, –ve or zero.

(ii) A line making an intercept of –a withy-axis means the line passing through (0,–a)

(iii) A line makes equal non-zero intercept with both co-ordinate axes then slope is –1

(iv) A line makes non-zero intercept with both co-ordinate axes equal in magnitude then slope is ±1.

The document Area of Triangle | Mathematics for NDA is a part of the NDA Course Mathematics for NDA.
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## Mathematics for NDA

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## FAQs on Area of Triangle - Mathematics for NDA

 1. What is the formula to find the area of a triangle?
Ans. The formula to find the area of a triangle is (base x height) / 2.
 2. How do you find the height of a triangle if it is not given?
Ans. If the height of a triangle is not given, you can use different methods depending on the information available. One approach is to use trigonometry, where you can use the length of one side and the measure of the angle opposite to that side to find the height. Another approach is to use the Pythagorean theorem if you have the lengths of all three sides of the triangle.
 3. Can you find the area of a triangle if only the lengths of the sides are given?
Ans. Yes, you can find the area of a triangle if the lengths of the sides are given. You can use Heron's formula, which states that the area of a triangle can be calculated using the lengths of its sides. The formula is √(s(s-a)(s-b)(s-c)), where s is the semi-perimeter of the triangle and a, b, and c are the lengths of its sides.
 4. Is it possible to find the area of a triangle if only the angles are given?
Ans. No, it is not possible to find the area of a triangle if only the angles are given. To calculate the area of a triangle, you need at least one side length or the length of the altitude (height) of the triangle.
 5. How do you find the area of an equilateral triangle?
Ans. To find the area of an equilateral triangle, you can use the formula (side length^2 * √3) / 4. In this formula, the side length represents the length of any side of the equilateral triangle.

## Mathematics for NDA

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