Answer:
None of the graphs of the equations pass through M(−10, 15).
Step-by-step explanation:
We are given the function [tex]y=kx[/tex].
It is given that the point B lies on the graph of the given function.
1. B(2, −3)
That is, x= 2 and y= -3
So, y= kx ⇒ -3 = 2k ⇒ [tex]k=\frac{-2}{3}[/tex]
Thus, the function is [tex]y=\frac{-2x}{3}[/tex]
2. [tex]B(3\frac{1}{3},-2)[/tex] i.e. [tex]B(\frac{10}{3},-2)[/tex]
That is, [tex]x= \frac{10}{3}[/tex] and y= -2
So, y= kx ⇒ [tex]-2 = \frac{10k}{3}[/tex] ⇒ [tex]k=\frac{-6}{10}[/tex] ⇒ [tex]k=\frac{-3}{5}[/tex]
Thus, the function is [tex]y=\frac{-3x}{5}[/tex]
From the graph plotted below, we get that,
None of the graphs of the equations pass through M(−10, 15).
