Answer:
The graph with B(2,-3) i.e) y=[tex]\frac{-3}{2}[/tex] x goes through the point M(-10,15).
Step-by-step explanation:
Consider M(-10,15) and given that equation is y = kx.
Now, substitute M(-10,15) in the equation
β 15 = k Γ -10
β k = [tex]\frac{15}{-10}[/tex] = [tex]\frac{-3}{2}[/tex]
β y = [tex]\frac{-3}{2}[/tex] x
Now, check with the given points B(2,-3) and B([tex]3\frac{1}{3}[/tex] , -2)
1) B(2,-3)
y = [tex]\frac{-3}{2}[/tex] x
β(-3) = [tex]\frac{-3}{2}[/tex] Γ 2
β -3 = -3 β LHS = RHS
β B(2,-3) is the required point.
2) for b([tex]3\frac{1}{3}[/tex] , -2)
LHS β RHS.
So,The graph with B(2,-3) i.e) y=[tex]\frac{-3}{2}[/tex] x goes through the point M(-10,15).
