Pearson Algebra 1 Common Core, 2011
PA
Pearson Algebra 1 Common Core, 2011 View details
6. Systems of Linear Inequalities
Continue to next subchapter

Exercise 42 Page 405

a In order to find the value of we first have to find the slope of the line This will allow us to find the slope of a line perpendicular to the given one. The given equation is in the standard form, so we will rearrange it so that it is written in slope-intercept form first.
In this equation the coefficient of the variable gives the slope of the line, Perpendicular lines have negative reciprocals for their slopes. In this case, this will be Since we can calculate the slope using the two given points on the line, and we can use the Slope Formula to solve for the missing value.
Solve for
b Now, having the slope and a point on the line, we can write a new equation using the point-slope form. We have found that the slope is or and we can use the point so and
Write in slope-intercept form
The equation of the line perpendicular to the given line and through the given points is
c To find a point that is a solution to both equations, we need to find their point of intersection. One way to do this is by equating the lines and solving for
Solve for
The coordinate of the common point is Now we can substitute this value of into one of the equations to find the corresponding coordinate.
The coordinate of the common point is . Therefore, the ordered pair is