Question：

Ask and be perpendicular to the straight line of $y = x -1$ and lead the straight line equation that cross a point {4,3}.

My method is:

{eqn = y == x - 1, pt = {4, 3}}
eq = # == 0 &@(eq /. Equal -> Subtract)
line = Cross@(Coefficient[eq[[1]], #] & /@ {x, y}) . {x, y} + c == 0
sol = First@(line /. Thread[{x, y} -> pt] // Solve)
line = line /. sol // Simplify // SubtractSides

Are there other methods to solve this problem?

Question：

Ask and be perpendicular to the straight line of $y = x -1$ and lead the straight line equation that cross a point {4,3}.

My method is:

{eqn = y == x - 1, pt = {4, 3}}
eq = # == 0 &@(eq /. Equal -> Subtract)
line = Cross@(Coefficient[eq[[1]], #] & /@ {x, y}) . {x, y} + c == 0
sol = First@(line /. Thread[{x, y} -> pt] // Solve)
line = line /. sol // Simplify // SubtractSides

This point may not necessarily lie on the original line. For example, a line that is perpendicular to the line y == x - 1 and passes through the point {6, 9}.

-15 + x + y == 0

Are there other methods to solve this problem?

Question：

Ask and be perpendicular to the straight line of $y = x -1$ and lead the straight line equation that cross a point {4,3}.

My method is:

{eqn = y == x - 1, pt = {4, 3}}
eq = # == 0 &@(eq /. Equal -> Subtract)
line = Cross@(Coefficient[eq[[1]], #] & /@ {x, y}) . {x, y} + c == 0
sol = First@(line /. Thread[{x, y} -> pt] // Solve)
line = line /. sol // Simplify // SubtractSides

Are there other methods to solve this problem?

Given the line equation ( y = x - 1 ), we need to find the equation of a line that is perpendicular to this line and passes through the point ((4, 3)).

First, rewrite the given line equation in standard form:

[ y - x + 1 = 0 ]

Next, identify the normal vector for this line. For a line in the form ( ax + by + c = 0 ), its normal vector is ((a, b)). Therefore, for the line ( y - x + 1 = 0 ), the normal vector is ((-1, 1)).

The slope of a line perpendicular to this one will be the negative reciprocal of the original line's slope. The slope of the original line is 1, so the slope of the perpendicular line is -1. Thus, a normal vector for the perpendicular line could be taken as ((1, 1)).

We can construct an equation of the form ( Ax + By + C = 0 ) where ((A, B)) is the normal vector of the perpendicular line. Here, ( A = 1 ) and ( B = 1 ), so we have:

[ x + y + C = 0 ]

Since this line passes through the point ((4, 3)), we can substitute the coordinates of this point into the above equation to solve for ( C ):

[ 4 + 3 + C = 0 ]

Solving for ( C ) gives us:

[ C = -7 ]

Therefore, the equation of the line that passes through the point ((4, 3)) and is perpendicular to the line ( y = x - 1 ) is:

[ x + y - 7 = 0 ]

Or written as:

[ y = -x + 7 ]

This is the equation of the required line.

Question：

Ask and be perpendicular to the straight line of $y = x -1$ and lead the straight line equation that cross a point {4,3}.

My method is:

{eqn = y == x - 1, pt = {4, 3}}
eq = # == 0 &@(eq /. Equal -> Subtract)
line = Cross@(Coefficient[eq[[1]], #] & /@ {x, y}) . {x, y} + c == 0
sol = First@(line /. Thread[{x, y} -> pt] // Solve)
line = line /. sol // Simplify // SubtractSides

Are there other methods to solve this problem?