Answer :

[tex]k:y=m_1x+b_1;\ l:y=m_2x+b_2\\\\k\ \perp\ l\iff m_1m_2=-1\\====================================\\k:y=-x-2;\ l:y=mx+b\\\\k\ \perp\ l\iff-1\cdot m=-1\Rightarrow m=1\\\\l:y=1x+b\to y=x+b\\\\(4;\ 1)\to substitute\ x=4\ and\ y=1\ to\ y=x+b:\\\\4+b=1\\b=1-4\\b=-3\\\\Answer:y=x-3[/tex]
AL2006
Hey !  I'll try.

You haven't said what the question is, and it doesn't show in the picture.

I'll assume that you're asked to find the equation of the line perpendicular to
the given line and passing through the point (4, 1).  Hope that's it.
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First, you need to remember that perpendicular lines have negative reciprocal slopes.

The given line has slope of -1, so the line perpendicular to it has slope of +1/1 = 1 .

So the equation of the new line is going to be [  y = 1x + intercept  ].

You know that the new line goes through the point where x=4 and y=1.
Stick these into the part of the equation that you already know:

y = 1x + intercept

1 = 1(4) + intercept.

Can you find the intercept from here ?
Maybe I'd better just finish it off.

1 = 4 + intercept.

Subtract 4 from each side:

intercept = -3

So the equation of the line perpendicular to [  y = -x - 2  ]  going through (4, 1) is

y = x - 3

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