The points $(-4,0),(4,0)$ and $(0,3)$ are the vertices of a\

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(b) : The points be $A(-4,0), B(4,0)$ and $C(0,3)$. Using distance formula $d=\sqrt{\left(x_{2}-x_{1}ight)^{2}+\left(y_{2}-y_{1}ight)^{2}}$ $$ \begin{aligned} & A B=\sqrt{(4-(-4))^{2}+(0-0)^{2}}=\sqrt{(4+4)^{2}}=\sqrt{8^{2}}=8 	ext { units } \ & B C=\sqrt{(0-4)^{2}+(3-0)^{2}}=\sqrt{16+9}=\sqrt{25}=5 	ext { units } \ & C A=\sqrt{(-4-0)^{2}+(0-3)^{2}}=\sqrt{16+9}=\sqrt{25}=5 	ext { units } \end{aligned} $$ And, $A B^{2} 
eq B C^{2}+C A^{2} \quad \because B C=C A$\ $	herefore \quad 	riangle A B C$ is an isosceles triangle.\