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# Algebra 2 Summer Review

Write an equation fort he line through the point (-1,2) and parallel to the line with equation 3x-2y-5=0. Write the equation in the general form.

We know that parallel lines have the same slope. Therefore, the line we are drawing will have the same slope as the equation 3x - 2y - 5 = 0.

However, to find the slope the equation must be in the form of: y = mx + b, where m = slope and b = y intercept.

3x - 2y - 5 = 0

2y = 3x - 5

y = (3/2)x - (5/2)

So, the slope of the line is: m = 3/2

We now know two pieces of information about the line, the slope is 3/2 and it goes through the point (-1, 2). Use the equation for point slope to generate the equation of our line:

(y - y1) = m (x - x1),

where m = 3/2, x1 = -1, y1 = 2

(y - 2) = (3/2) (x - (-1))

y - 2 = (3/2)x + (3/2)

y = (3/2)x + (7/2)

Answer: y = (3/2)x + (7/2)

If the line is given in the form ax+by+c=0 then coefficients a and b give the direction perpendicular to the line. If you know vectors, those are the coordinates of a vector directed perpendicular to the line. Anyway, we know from geometry that a line perpendicular to one of the two parallel lines is perpendicular to another one too. So the line parallel to the given one 3x-2y-5=0 has to have the same coefficients a and b because the direction perpendicular to it is the same direction, which is perpendicular to the line 3x-2y-5=0. So it is in the forms 3x-2y+c=0. The c coefficient is found by plugging in the coordinates of a point, through which the line passes, that is (-1,2). The result is: 3*(-1)-2*2+c=0 or -7+c=0. It is easy to see now that c=7. The line equation is thus 3x-2y+7=0.