Question

Find the area of triangle formed by the points (a,1/b), (b,1/b), (c,1/c)​

Answer 1

Answer:

Step-by-step explanation:

1/2[y1{x2-x3} +y2{x3-x1} +y3{x1-x2}]

1/2[1/b{b-c} + 1/b{c-a} + 1/c{a-b}]

1/2[1-c/b + c/b-a/b +a/c - b/c]

1/2[1-a/b + a/c -b/c]

this is the area of triangle

Answer 2

Concept: As the study of geometry utilising coordinate points, coordinate geometry is a term. In coordinate geometry, a triangle's three vertices can be specified in the coordinate plane to determine the triangle's area. In coordinate geometry, the region or space a triangle occupies in the 2-D coordinate plane is known as its area.

Given: points (a,1/b), (b,1/b), (c,1/c)​

To find: the area of triangle

Solution: to find the area of the triangle, we use the formula

$area = |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

Here,

$x_1 = a\\x_2 = b\\x_3 = c$

and

$y_1 = \frac{1}{b} \\\\y_2 = \frac{1}{b} \\\\y_3 = \frac{1}{c}$

Therefore, area of triangle =

$|a(\frac{1}{b} - \frac{1}{c}) + b(\frac{1}{c} - \frac{1}{b}) + c(\frac{1}{b} - \frac{1}{b})|\\\\|a(\frac{c - b}{bc}) + b(\frac{b-c}{bc}) + c(\frac{b-b}{b^{2} })|\\\\|a(\frac{c - b}{bc}) + b(\frac{b-c}{bc}) + 0)|\\\\|(\frac{ac-ab}{bc} ) + (\frac{b^2-bc}{bc})|\\\\|\frac{ac-ab+b^2-bc}{bc}|$

$\frac{ac-ab+b^2-bc}{bc}$

Hence, the area of the triangle formed by the points (a,1/b), (b,1/b), (c,1/c)​ is $\frac{ac-ab+b^2-bc}{bc}$ sq units.

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