Question

help?

Answer 1

help me

Answer 2

r u still? helping?

Answer 3

\[y-y_1=m(x-x_1)\]
M= slope \[x_1,y_1\]= Point of line


So were' given y=-2x-7

With using y=mx+b

We have \[m_1=-2\]

We are being told that there's a new path being taken. (Perpiniculer)


\[m_1 \times m_2=-1\]

Input what we have:

\[-2 \times m_2=-1\]

\[=> m_2=\frac{ 1 }{ 2 }\]

m2= slope of new path

New path intersects with (-2,-3)

Input o;

\[y-(-3)=\frac{ 1 }{ 2 } (x-(-2))\]

\[=> y+3=\frac{ 1 }{ 2 } (x+2)\]

\[=>y+3=\frac{ 1 }{ 2 }x+1\]

Subtract 3 from each side
\[y=\frac{ 1 }{ 2 }x-2\]

So forth, \[y=\frac{ 1 }{ 2 }x-2\]
Should be your answer for the new path.

Answer 4

*perpendicular

Answer 5

tysm