Prove that

Solution:

(i) Let $A\left(x_{1}, y_{1}\right)=A(1,-3), B\left(x_{2}, y_{2}\right)=B(4, p)$ and $C\left(x_{3}, y_{3}\right)=C(-9,7)$. Now

Area $(\Delta A B C)=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]$

$\Rightarrow 15=\frac{1}{2}[1(p-7)+4(7+3)-9(-3-p)]$

$\Rightarrow 15=\frac{1}{2}[10 p+60]$

$\Rightarrow|10 p+60|=30$

Therefore

$\Rightarrow 10 p+60=-30$ or 30

$\Rightarrow 10 p=-90$ or $-30$

$\Rightarrow p=-9$ or $-3$

Hence, $p=-9$ or $p=-3$.

(ii)

Let $A\left(x_{1}, y_{1}\right)=A(2,1), B\left(x_{2}, y_{2}\right)=B(3,-2)$ and $C\left(x_{3}, y_{3}\right)=C\left(\frac{7}{2}, y\right)$.

Now

Area $(\Delta A B C)=\frac{1}{2}\left|x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right|$

$\Rightarrow 5=\frac{1}{2}\left|2(-2-y)+3(y-1)+\frac{7}{2}(1+2)\right|$

$\Rightarrow 10=\left|-4-2 y+3 y-3+\frac{21}{2}\right|$

$\Rightarrow 10=\left|y+\frac{7}{2}\right|$

$\Rightarrow 10=y+\frac{7}{2} \quad$ or $-10=y+\frac{7}{2}$

$\Rightarrow y=\frac{13}{2}$ or $y=\frac{-27}{2}$

Hence, $y=\frac{13}{2}$ or $\frac{-27}{2}$.