Let's find the y-intercept of the line with equation
![-7x+6y=-3](https://img.qammunity.org/2023/formulas/mathematics/college/tktbf4mms9cwv7zyhj0atnutl7vcs37orf.png)
To get the y-intercept, we plug in "0" into "x". So,
![\begin{gathered} -7x+6y=-3 \\ -7(0)+6y=-3 \\ 6y=-3 \\ y=(-3)/(6) \\ y=-(1)/(2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/frnjjluxbtonxbfv3rvh0ibhp2829mzusf.png)
So, the coordinate point is
![(0,-(1)/(2))](https://img.qammunity.org/2023/formulas/mathematics/college/ywymdz1u6c2jfcb7wj9rzipk5bq0e79x15.png)
We need to find the equation of the line that is perpendicular to the line with equation -2x + 4y = 2 and passes through the point (0, -1/2).
First, let's re-arrange -2x + 4y = 2 into the form y = mx + b, where m is the slope and b is the y-intercept.
![\begin{gathered} -2x+4y=2 \\ 4y=2x+2 \\ y=(2x)/(4)+(2)/(4) \\ y=(1)/(2)x+(1)/(2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/wjmo7zp8oh1i7028ft0f9r98hmpobmxp0l.png)
We know the line perpendicular will have a slope that is negative reciprocal.
So, the perpendicular line will have a slope of
![-2](https://img.qammunity.org/2023/formulas/mathematics/high-school/6r32gfn59zzqnd2b7wfav8dfc8jefwu0gf.png)
So, the line will take the form:
![y=-2x+b](https://img.qammunity.org/2023/formulas/mathematics/high-school/rlu6nfc7816wamuzwxl9pfle23oh7bwf7c.png)
Since it goes through the point (0, -1/2), we can solve for "b":
![\begin{gathered} y=-2x+b \\ -(1)/(2)=-2(0)+b \\ -(1)/(2)=b \\ \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/h6naz75qvhf0jnq9rik54r02n8pi075y1d.png)
Thus, the equation of the linne is,
![y=-2x-(1)/(2)](https://img.qammunity.org/2023/formulas/mathematics/college/6ij0v6vz38au9vs9k5xoi5ir6evxqwzy4w.png)