Question

Find the area of the triangle whose vertices are ${(at_1^2,2a{t_1}),(at_2^2,2a{t_2}){\text{ and }}(at_3^2,2a{t_3})}$

Answer 1

Hint:  Just remember the formula for the area of the triangle, and then by simply putting the values you will get the answer.

Complete step by step answer:

We know that,

The area of a triangle = $\dfrac{1}{2}[{x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2})$

Here we have,

$\dfrac{1}{2}[at_1^2(2a({t_2} - {t_3})) + at_2^2(2a({t_3} - {t_1})) + at_3^2(2a({t_1} - {t_2})]$

$\Rightarrow$ $\dfrac{1}{2}[at_1^2(2a{t_2} - 2a{t_3}) + at_2^2(2a{t_3} - 2a{t_1}) + at_3^2(2a{t_1} - 2a{t_2})]$

Taking $2{a^2}$ as common, we get,

$\dfrac{1}{2}(2{a^2})[t_1^2({t_2} - {t_3}) + t_2^2({t_3} - {t_1}) + t_3^2({t_1} - {t_2})]$

$\Rightarrow$ ${a^2}(t_1^2{t_2} - t_1^2{t_3} + t_2^2{t_3} - t_2^2{t_1} + t_3^2{t_1} - t_3^2{t_2})]$

$\Rightarrow$  ${a^2}[(t_1^2{t_2} - t_3^2{t_2}) + (t_2^2{t_3} - t_1^2{t_3}) + (t_3^2{t_1} - t_2^2{t_1})]$

$\Rightarrow$ ${a^2}[{t_2}(t_1^2 - t_3^2) + {t_3}(t_2^2 - t_1^2) + {t_1}(t_3^2 - t_2^2)]$

Now using identity ${a^2} - {b^2}$ = (a + b)(a - b)

${a^2}[{t_2}({t_1} + {t_3})({t_1} - {t_3}) + {t_3}({t_2} + {t_1})({t_2} - {t_1}) + {t_1}({t_3} + {t_2})({t_3} - {t_2})]$

So, this is the required answer.

Note: You should be careful while choosing the formula for finding the area for these kinds of questions. Since the vertices are given, we have to take the formula of area of the triangle in coordinate geometry.