The lengths of sides are equal, hence the triangle formed are congruent.
Step-by-step explanation:
With the given information we need to draw the graph first
draw x axis and y axis
a: y=3x+2
b: y=-1/3x-1
c: passes through points (1,5) and (3,-2)
d: passes through points (-6,1) and (-3,-7)
Using the points given draw the four lines and mark the points
we have four lines namely a,b,c,d
we have Δ AOB and Δ COD
AB = [tex]\sqrt{(3-1)^2 +(-2-5)^2}\\[/tex]
= [tex]\sqrt{53}[/tex]
CD = [tex]\sqrt{(-6+3)^2 +(1+7)^2}\\[/tex]
= [tex]\sqrt{75}[/tex]
equation of BD = x+3y +3 =0
equation of AC = y=3x+2
therefore coordinates of 0 ≡ ([tex]\frac{-3}{10}[/tex],[tex]\frac{-7}{10}[/tex] )
OD = 1.7 [tex]\sqrt{10}[/tex] OC = 2.1 [tex]\sqrt{10}[/tex]
OB = 1.3 [tex]\sqrt{10}[/tex] OA = 1.9 [tex]\sqrt{10}[/tex]
None of the corresponding lengths of sides are equal. hence the triangle formed are congruent.
