ALGEBRA

Nameinstitution

Unit

The equation I chosewas y= 2x – 1 (-2, 2)

First, I must workon the parallel line which must pass through point (-2, 2). Parallellines are lines that run side by side but can never meet and theyhave a similar slope.

Borrowingfrom the general form of thepoint-slope equation y– y1 =m(x – x1),I can calculate the equation of a line parallel to line y = 2x-1 andpassing through points (-2, 2). Thus, my new equation is:

y- 2 = 2 [x – (-2)]

y– 2 = 2 [x + 2]

y-2= 2x + 4

y=2x + 6

Liney =2x + 6 passes points (-2, 2) and is parallel to line y= 2x-1

Perpendicularline

Perpendicularlines are lines that intersect one another at an angle of 90^{o}.

Toderive an equation of a line perpendicular to the original line y= 2x– 1, and passing through ordered points (-2, 2), I shall employ thegeneral form of thepoint-slope equation, y – y1= m(x – x1).However, for perpendicular lines, the slope of one line is equal tothe negative reciprocal of the other line. Thus for my unknown line,the slope shall be –½

Thussubstituting the ordered points on the equation I have:

y -y1 =-½ [x- (-2)]

y-2 = -½(x+2)

y-2 = -½ x – 1

y= -½x+1

Thusthe y-intercept for this new perpendicular line is one unit above theorigin.

Tocalculate the x- intercept, I derive the equation for x.

y= -½x+1 ………multiply both sides by -2 to eliminate the fraction andthe negative on x

-2(y) =(-½ +1) -2

-2y = x– 2

x = -2y +2

Thus thex-intercept is 2 units to the left of the origin.