Question

the triangle abc is translated 1 units left and 2 units up. find the vertices of the triangle abc. a = (-4, 3) b = (-5, -7) c = (4, 3) a. a(-3, 5), b(-4, -5), c(3, 7) b. a(-5, 5), b(-6, -5), c(3, 5) c. a(2, -5), b(6, -1), c(2, 5) d. a(3, -5), b(-6, -3), c(-3, 3)

Explanation:

Step1: Determine the translation rule

Translation \(1\) unit left means subtract \(1\) from the \(x -\)coordinate. Translation \(2\) units up means add \(2\) to the \(y -\)coordinate. The general rule for a point \((x,y)\) is \((x - 1,y+2)\).

Step2: Apply the rule to point \(A(-4,3)\)

For \(A(-4,3)\):
\(x=-4,y = 3\)
\(x'=-4 - 1=-5\), \(y'=3 + 2=5\). So \(A'(-5,5)\)

Step3: Apply the rule to point \(B(5, - 7)\)

For \(B(5,-7)\):
\(x = 5,y=-7\)
\(x'=5-1 = 4\), \(y'=-7 + 2=-5\). So \(B'(4,-5)\)

Step4: Apply the rule to point \(C(-4,3)\)

For \(C(-4,3)\):
\(x=-4,y = 3\)
\(x'=-4-1=-5\), \(y'=3 + 2=5\). Wait, no, original \(C\) is \((-4,3)\)? Wait, no, looking at the right - hand side labels: \(A=(-4,3)\), \(B=(5,-7)\), \(C=(4,3)\)
For \(C(4,3)\):
\(x = 4,y = 3\)
\(x'=4-1=3\), \(y'=3 + 2=5\). So \(C'(3,5)\)

Answer:

B. \(A'(-5,5),B'(4,-5),C'(3,5)\)