# find an equation of a line through the given point and parallel to and perpendicular to y=2(x+1)-2 at (2,1)

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### 1 Answer

Given the line `y=2(x+1)-2` , find the equation of a line (i) parallel and (ii) perpendicular to the given line that goes through the point (2,1)

(1) Lines that are parallel have the same slope. The slope of the given line is 2. (This is the line y=2x shifted left one unit, and down two units)

Thus we want a line with slope 2 that contains (2,1). We use the point-slope form (`y-y_1=m(x-x_1)` where m is the slope and `(x_1,y_1)` is the given point)

`y-1=2(x-2)` is the equation we seek. Other forms are `y=2(x-2)+1` or `y=2x-3`

(2) Lines that are perpendicular have opposite-reciprocal slopes. Given the slope of a line `m_1` , the slope of the line perpendicular to it is `-1/m_1` (Or equivalently the product of the slopes is -1)

Thus the slope of the perpendicular line is `(-1)/2` . Again we use the point-slope form to get:

`y-1=(-1)/2(x-2)` or `y=(-1)/2(x-2)+1` or `y=(-1)/2x+2`