(100 points boys) Just this one question for my math.

Accepted Solution

Answer:[tex]A^{'}[/tex] = (1,[tex]\frac{5}{2}[/tex]), [tex]B^{'}[/tex] = ([tex]\frac{3}{2}[/tex],-1) and [tex]C^{'} = (2,2)[/tex]OA = [tex]\sqrt[]{29}[/tex] units, OB = [tex]\sqrt[]{13}[/tex] units and OC = [tex]\sqrt[]{32}[/tex] unitsO[tex]A^{'}[/tex] = [tex]\frac{\sqrt{29} }{2}[/tex] units,O[tex]B^{'}[/tex] = [tex]\frac{\sqrt{13} }{2}[/tex] units andO[tex]C^{'}[/tex] = [tex]\frac{\sqrt{32} }{2}[/tex] unitsStep-by-step explanation:Given vertices are A(2,5), B(3,-2), C(4,4) and the new vertices after dilation are [tex]A^{'}, B^{'}, C^{'}[/tex]a. Triangle needs to be dilated about the origin.β‡’The centre of dilation is the origin O(0,0).scaling factor is [tex]\frac{1}{2}[/tex][tex]\frac{OA^{'}}{OA}[/tex] = [tex]\frac{OB^{'}}{OB}[/tex] = [tex]\frac{OC^{'}}{OC}[/tex] = [tex]\frac{1}{2}[/tex]β‡’ [tex]A^{'}, B^{'} and C^{'}[/tex] divide OA, OB and OC in the ratio 1:1 respectively.Using the formula for the coordinates of a point dividing two points (0,0), (x,y) in the ratio 1:1 is ([tex]\frac{x}{2}[/tex],[tex]\frac{y}{2}[/tex]) :[tex]A^{'}[/tex] = (1,[tex]\frac{5}{2}[/tex]), [tex]B^{'}[/tex] = ([tex]\frac{3}{2}[/tex],-1) and [tex]C^{'} = (2,2)[/tex]b.Using the formula for the distance between the points (0,0) and (x,y) is [tex]\sqrt[]{x^{2 }+ y^{2} }[/tex] : OA = [tex]\sqrt[]{2^{2} +5^{2}}[/tex] = [tex]\sqrt[]{29}[/tex] unitsOB = [tex]\sqrt[]{3^{2} +-2^{2}}[/tex] = [tex]\sqrt[]{13}[/tex] unitsOC = [tex]\sqrt[]{4^{2} +4^{2}}[/tex] = [tex]\sqrt[]{32}[/tex] unitsc.From the property of dilation : [tex]\frac{OA^{'}}{OA}[/tex] = [tex]\frac{OB^{'}}{OB}[/tex] = [tex]\frac{OC^{'}}{OC}[/tex] = [tex]\frac{1}{2}[/tex]β‡’O[tex]A^{'}[/tex] = [tex]\frac{OA}{2}[/tex] = [tex]\frac{\sqrt{29} }{2}[/tex] unitsO[tex]B^{'}[/tex] = [tex]\frac{OB}{2}[/tex] = [tex]\frac{\sqrt{13} }{2}[/tex] unitsO[tex]C^{'}[/tex] = [tex]\frac{OC}{2}[/tex] = [tex]\frac{\sqrt{32} }{2}[/tex] unitsd. Using the formula for the distance between two points :O[tex]A^{'}[/tex] = [tex]\sqrt{1^{2} + (\frac{5}{2})^{2} }[/tex] = [tex]\frac{\sqrt{29} }{2}[/tex] unitsO[tex]B^{'}[/tex] = [tex]\sqrt{(\frac{3}{2})^{2} + -1^{2}}[/tex] = [tex]\frac{\sqrt{13} }{2}[/tex] unitsO[tex]C^{'}[/tex] = [tex]\sqrt[]{2^{2} +2^{2}}[/tex] = [tex]\frac{\sqrt{32} }{2}[/tex] unitswhich are the same as we predicted in part c.