Answer:
![y=-(3)/(2)x -7](https://img.qammunity.org/2020/formulas/mathematics/middle-school/4gc8kh3s1raqcehssz8y1w49y0t2ufcc9y.png)
Explanation:
Line L has the equation:
![2x-3y=5](https://img.qammunity.org/2020/formulas/mathematics/middle-school/lo7d54hetpk38youtfpy1dado4t8ys47mu.png)
we need to clear for y:
![-3y=-2x+5\\y=(2)/(3) x-(5)/(3)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/1yt8ty0qpgy8j1ip07u6lnkz6d7mqan8tn.png)
now we have the form a general line equation
where
is the slope of the line.
so the slope of the line L is:
![m_(1)=(2)/(3)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/rlgr5j2evb3umiqqfozpupk4n5x97hdjl4.png)
and for two lines to be parallel the following condition must be met
![m_(1)*m_(2)=-1](https://img.qammunity.org/2020/formulas/mathematics/middle-school/b4wzgvi0m29ljdgiewrn0t12yhydcouoeb.png)
where
in this case is the slope of line M, substituting the value
to find
:
![(2)/(3)*m_(2)=-1\\ m_(2)=(-1(3))/(2)\\m_(2)=-(3)/(2)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/cyh11m0uxd8ybdlkdutqgasyv6atp364ux.png)
This is the slope of line M, and since we also know that it passes through the point (2, -10) where
and
![y_(0)=-10](https://img.qammunity.org/2020/formulas/mathematics/middle-school/82upzv307xhiuwh2s0waujdyi0lh3dz049.png)
we use the point- slope equation and substitute known values to find the equation of the line M:
![y-y_(0)=m(x-x_(0))\\y-(-10)=-(3)/(2)(x-2)\\ y+10=-(3)/(2)x+3\\y=-(3)/(2)x +3-10\\y=-(3)/(2)x -7](https://img.qammunity.org/2020/formulas/mathematics/middle-school/v7huxvuvfp0184jqj1numrs0r4ekwm3mcs.png)
the equation of line M is:
![y=-(3)/(2)x -7](https://img.qammunity.org/2020/formulas/mathematics/middle-school/4gc8kh3s1raqcehssz8y1w49y0t2ufcc9y.png)