Write an equation of the line that passes through the given point and is (a) parallel and (b) perpendicular to the given line (in the description).

Question

Answer 1

The question doesn't specify a form for the equation, so I'll use point-slope form since it is the most natural here. As a reminder, the form looks like:

y - y₁ = m(x - x₁)

where x₁ and y₁ refer to any point on the line. Given the point (3, -2), our equation becomes:

y - (-2) = m(x - 3) or y + 2 = m(x - 3).

The reference line seems to pass through the points (1, -4) and (2, -1) which gives a slope of 3. So, the parallel line is y + 2 = 3(x - 3). The perpendicular line's slope is the negative of the reciprocal, or -1/3. Finally, our perpendicular line is described by y + 2 = -1/3 * (x - 3).

Answer 2

two points on given line: (2,-1) and (1,-4)

slope = (-4--1)/(1-2) = 3

slope of line parallel to that line will have equal slope

(3,-2) is point on the line.

equation of a line: y = mx + b . need to find m and b, where m is slope and b is y-intercept

we have determined m = 3, to find b, substitute point (x,y) = (3,-2) for x and y in equation:

-2 = 3*3 + b

-11 = b

y = 3x - 11 is equation of line parallel.

slope of line perpendicular to that line will have slope equal to negative reciprocal of 3 = -1/3

again substitute point (x,y) = (3,-2) for x and y in equation where m = -1/3:

-2 = -1/3*3 + b

-2 = -1 + b

-1 = b

y = -1/3x -1 is equation of line perpendicular to the given line