Question

$△ABC$ with vertices $A(2,a),B(3,b)\&C(3,c)$ translates $4$ units down and gets translated to $△A^{\prime }{B}^{\prime }{C}^{\prime }$ with vertices $A^{\prime }(2,-2),B^{\prime }(3,-1)\&C^{\prime }(3,-3)$. The value of $a+b+c$ is

Answer 1

Given : $△ABC$ with vertices $A(2,a),B(3,b)\&C(3,c)$ translates $4$ units down and gets translated to $△A^{\prime }{B}^{\prime }{C}^{\prime }$ with vertices $A^{\prime }(2,-2),B^{\prime }(3,-1)\&C^{\prime }(3,-3)$

Solution:
Vertex $A(2,a)$ shifts $4$ units down to $A^{\prime }(2,-2)$ that means $a-4=-2\Rightarrow a=2$

Vertex $B(3,b)$ shifts $4$ units down to $B^{\prime }(3,-1)$ that means $b-4=-1\Rightarrow b=3$

Vertex $C(3,c)$ shifts $4$ units down to $C^{\prime }(3,-3)$ that means $c-4=-3\Rightarrow c=1$

$\therefore a+b+c=2+3+1=6$