Find the area of a triangle whose vertices are

Solution:

We know area of triangle formed by three points $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ is given by $\Delta=\frac{1}{2}\left|x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right|$

(i) The vertices are given as $(6,3),(-3,5),(4,-2)$.

$\Delta=\frac{1}{2}|6(5+2)-3(-2-3)+4(3-5)|$

$=\frac{1}{2}|6 \times 7-3 \times(-5)+4 \times(-2)|$

$=\frac{1}{2}|42+15-8|$

$=\frac{49}{2}$ sq units

(ii) The vertices are given as $\left(a t_{1}^{2}, 2 a t_{1}\right),\left(a t_{2}^{2}, 2 a t_{2}\right),\left(a t_{3}^{2}, 2 a t_{3}\right)$.

$\Delta=\frac{1}{2}\left|a t_{1}^{2}\left(2 a t_{2}-2 a t_{3}\right)+a t_{2}^{2}\left(2 a t_{3}-2 a t_{1}\right)+a t_{3}^{2}\left(2 a t_{1}-2 a t_{2}\right)\right|$

$=\frac{1}{2} \cdot 2 a^{2}\left|\left(t_{1}{ }^{2} t_{2}-t_{1}{ }^{2} t_{3}\right)+\left(t_{2}{ }^{2} t_{3}-t_{2}{ }^{2} t_{1}\right)+\left(t_{3}{ }^{2} t_{1}-t_{3}{ }^{2} t_{2}\right)\right|$

$=a^{2}\left|\left(t_{1}{ }^{2} t_{2}-t_{2}{ }^{2} t_{1}\right)+\left(t_{2}{ }^{2} t_{3}-t_{1}{ }^{2} t_{3}\right)+\left(t_{3}{ }^{2} t_{1}-t_{3}{ }^{2} t_{2}\right)\right|$

$=a^{2}\left|t_{1} t_{2}\left(t_{1}-t_{2}\right)+t_{3}\left(t_{2}^{2}-t_{1}^{2}\right)+t_{3}^{2}\left(t_{1}-t_{2}\right)\right|$

$=a^{2}\left|\left(t_{1}-t_{2}\right)\left\{t_{1} t_{2}-t_{3}\left(t_{2}+t_{1}\right)+t_{3}{ }^{2}\right\}\right|$

$=a^{2}\left|\left(t_{1}-t_{2}\right)\left\{t_{1} t_{2}-t_{3}\left(t_{2}+t_{1}\right)+t_{3}^{2}\right\}\right|$

$=a^{2}\left|\left(t_{1}-t_{2}\right)\left\{t_{1} t_{2}-t_{3} t_{2}-t_{3} t_{1}+t_{3}^{2}\right\}\right|$

$=a^{2}\left|\left(t_{1}-t_{2}\right)\left\{t_{2}\left(t_{1}-t_{3}\right)-t_{3}\left(-t_{3}+t_{1}\right)\right\}\right|$

$=a^{2}\left|\left(t_{1}-t_{2}\right)\left(t_{1}-t_{3}\right)\left(t_{2}-t_{3}\right)\right|$

or, $\Delta=a^{2}\left(t_{1}-t_{2}\right)\left(t_{2}-t_{3}\right)\left(t_{3}-t_{1}\right)$ assuming $t_{1}>t_{2}, t_{2}>t_{3}, t_{3}>t_{1}$

(iii)

The vertices are given as $(a, c+a),(a, c),(-a, c-a)$.

$\Delta=\frac{1}{2}|a(c-c+a)+a(c-a-c-a)-a(c+a-c)|$

$=\frac{1}{2}|a(a)+a(-2 a)-a(a)|$

$=\frac{1}{2}\left|-2 a^{2}\right|=a^{2}$