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
⇒ 15 = k × -10
⇒ k = [tex]\frac{15}{-10}[/tex] = [tex]\frac{-3}{2}[/tex]
⇒ 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
⇒(-3) = [tex]\frac{-3}{2}[/tex] × 2
⇒ -3 = -3 ⇒ LHS = RHS
⇒ B(2,-3) is the required point.
2) for b([tex]3\frac{1}{3}[/tex] , -2)
LHS ≠ RHS.
So,The graph with B(2,-3) i.e) y=[tex]\frac{-3}{2}[/tex] x goes through the point M(-10,15).
