Chapter Section : Equations of Lines Answers to Problems For problems -, put our answers into slope intercept form..* Find the equation of the line with slope, and passing through the point (,0).. Find the equation of the line with slope, and passing through the point (, ). = x + = x.* Find the equation of the line with slope,.* Find the equation of the line with slope, and passing through the point (0,). = x + and passing through the point (, ). = x.* Find the equation of the line with slope, and passing through the point (, ). = x 6. Find the equation of the line with slope, and passing through the point (, 0). = x + 9 7.* Find the equation of the line with slope, 8. Find the equation of the line with slope 7, and passing through the point (, ). = x and passing through the point (, ). 7 7 = x 9.* Find the equation of the line with slope, and passing through the point (, ). 0. Find the equation of the line with slope 0, and passing through the point (, ). = x + =.* Find the equation of the line with undefined slope, and passing through the point (, ). x =. Find the equation of the line with slope, and passing through the point (, ). = x *See Explanations & Worked Solutions on page. Page of
Chapter Section : Equations of Lines Find the equation of the line passing through the given points, for problems -0..* (0,) and (,). (, ) and (, ) = x + =.* (,) and (,) 6. (, ) and (,) = x + = x 7.* (, ) and (, ) 8. (,) and (, ) = x x = 9.* (0,) and (,0) 0. (,) and (, 6) = x + 7 = x 6.* Find the equation of the line parallel to x = and passing through (, ). 8 = x.* Find the equation of the line parallel to x + = and passing through (0,0). = x.* Find the equation of the line parallel to = 6x and passing through (,).. Find the equation of the line parallel to = x + and passing through (,). = x. Find the equation of the line parallel to x + = and passing through (,). 0 = x + 6. Find the equation of the line parallel to 6 x = and passing through (, ). = 6x x = 7.* Find the equation of the line perpendicular to x = 0 and passing through (, ). = x 9.* Find the equation of the line perpendicular to = x and passing through (,). = x + 8. Find the equation of the line perpendicular to x = and passing through (, ). = x 6 0. Find the equation of the line perpendicular to x = and passing through (, 6). 8 = x *See Explanations & Worked Solutions on page. Page of
Chapter Section : Equations of Lines.* Find the equation of the line perpendicular to = x + and passing through (, ). = x +. Find the equation of the line perpendicular to = x and passing through (, ). = x.* Graph the line = x + using onl the line s slope and intercept.. Graph the line = x using onl the line s slope and intercept..* Graph the line x + = 6 using onl the line s x and intercepts. 6. Graph the line x + = using onl the line s x and intercepts. 7.* Graph the line x + = 6 using onl the line s x and intercepts. *See Explanations & Worked Solutions on page. Page of
Chapter Section : Equations of Lines Find the equation of the graphed line, for problems 8-0. 8. 9.* = x = x + 0. = *See Explanations & Worked Solutions on page. Page of
Chapter Section : Equations of Lines *Explanations & Worked Solutions. Start with the slope-intercept form for the equation of the line, with - in place of m. Then substitute the given point s x and -coordinates into the equation and solve for the -intercept, b: = x (, 0) ( ) 0 = 0 = + + So, the line s equation is = x +.. Start with the slope-intercept form for the equation of the line, with in place of m. Then substitute the given point s x and -coordinates into the equation and solve for the -intercept, b: = x ( 0, ) = ( 0) So, the line s equation is = x +.. Start with the slope-intercept form for the equation of the line, with / in place of m. Then substitute the given point s x and -coordinates into the equation and solve for the -intercept, b: = x, (, ) = ( ) = + + + + So, the line s equation is = x. Page of
Chapter Section : Equations of Lines. Start with the slope-intercept form for the equation of the line, with in place of m. Then substitute the given point s x and -coordinates into the equation and solve for the -intercept, b: = x (, ) = ( ) = So, the line s equation is = x. 7. Start with the slope-intercept form for the equation of the line, with in place of m. Then substitute the given point s x and -coordinates into the equation and solve for the -intercept, b: = x (, ) = ( ) = 0 So, the line s equation is = x. 9. Start with the slope-intercept form for the equation of the line, with in place of m. Then substitute the given point s x and -coordinates into the equation and solve for the -intercept, b: = x (, ) ( ) = = 9 So, the line s equation is = x +.. Onl vertical lines have undefined slope, so this line is vertical. Since it passes through the point (, ), it must have equation x =.. First we find the line s slope m, the ratio of the difference of the -coordinates to the difference of the x- coordinates: m = = =. Now, use the slope-intercept form for the equation of the line, with 0 in place of m. continued Page 6 of
Chapter Section : Equations of Lines Then substitute the x and -coordinates of either of the given points into the equation and solve for the - intercept, b. We ll use the point (0,): = x, 0, So, the equation of the line is = x+. ( 0) = ( ). First we find the line s slope m, the ratio of the difference of the -coordinates to the difference of the x- coordinates: m = = =. Now, use the slope-intercept form for the equation of the line, with / in place of m. Then substitute the x and -coordinates of either of the given points into the equation and solve for the -intercept, b. We ll use the point (,): = x, (, ) = ( ) = So, the line s equation is = x +. 7. First we find the line s slope m, the ratio of the difference of the -coordinates to the difference of the x- ( ) ( ) + coordinates: m = = = =. Now, use the slope-intercept form for the equation ( ) ( ) + of the line, with in place of m. Then substitute the x and -coordinates of either of the given points into the equation and solve for the -intercept, b. We ll use the point (, ): = x,, So, the equation of the line is = x. = = + + ( ) ( ) 9. (0,) and (,0) First we find the line s slope m, the ratio of the difference of the -coordinates to the difference of the x- coordinates: m = 0 = 0 =. Now, use the slope-intercept form for the equation of the line, with in place of m. continued Page 7 of
Chapter Section : Equations of Lines Then substitute the x and -coordinates of either of the given points into the equation and solve for the - intercept, b. We ll use the point (0,): = x, 0, So, the equation of the line is = x+. ( 0) = ( ). Parallel lines have equal slopes, so our line has the same slope as the line x =. To see what that slope is, put the line into slope-intercept form (i.e., solve for ). x = x x = x + x = + = x So the line we seek also has a slope m = /. Now use the slope-intercept form for the equation of the line, with / in place of m. Then substitute the x and -coordinates of either of the given points into the equation and solve for the -intercept, b. We ll use the point (, ): = x, (, ) = ( ) = + + + + 8 8 So, the line s equation is = x.. Parallel lines have equal slopes, so our line has the same slope as the line x + =. To see what that slope is, put the line into slope-intercept form (i.e., solve for ). continued Page 8 of
Chapter Section : Equations of Lines x + = x x = x + x = + = x + So the line we seek also has a slope m = -/. Now use the slope-intercept form for the equation of the line, with -/ in place of m. Then substitute the x and -coordinates of either of the given points into the equation and solve for the -intercept, b. We ll use the point (0,0): = x, ( 0, 0) 0 = ( 0) 0 So, the equation of the line is = x.. Parallel lines have equal slopes, so our line has the same slope as the line = 6x. So the line we seek also has a slope m = 6. Now use the slope-intercept form for the equation of the line, with 6 in place of m. Then substitute the x and -coordinates of either of the given points into the equation and solve for the -intercept, b. We ll use the point (, ) : = 6 x,, ( ) = 6 = 6+ b 6 6 ( ) So, the equation of the line is = 6x. 7. Perpendicular lines have reciprocal slopes. So if we find the slope of the given line, its reciprocal will be the slope of the line that we seek. To see what that slope is, put the line into slope-intercept form (i.e., solve for ). x = 0 x x = x + 0 x 0 = + = x So the line we seek also has slope m = /. Now use the slope-intercept form for the equation of the line, with / in place of m. continued Page 9 of
Chapter Section : Equations of Lines Then substitute the x and -coordinates of either of the given points into the equation and solve for the - intercept, b. We ll use the point (, ): = x, (, ) = ( ) = 0 So, the line s equation is 0 0 = x. 9. Perpendicular lines have reciprocal slopes. So the line we seek also has slope m =. Now use the slopeintercept form for the equation of the line, with in place of m. Then substitute the x and -coordinates of either of the given points into the equation and solve for the -intercept, b. We ll use the point (,): = x,, So, the line s equation is = x +. ( ) = = + + ( ). Find the equation of the line perpendicular to = x + and passing through (, ). Perpendicular lines have reciprocal slopes. So the line we seek also has slope m =. Now use the slope- intercept form for the equation of the line, with in place of m. Then substitute the x and -coordinates of either of the given points into the equation and solve for the -intercept, b. We ll use the point (, ): = x, (, ) = ( ) = + + So, the line s equation is = x +. Page 0 of
Chapter Section : Equations of Lines. The x-intercept occurs when = 0, and the -intercept occurs when x = 0. When = 0 we have: 0 = x + = x = x So the x-intercept is x =. When x = 0 we have: = 0 + = 0+ = So the -intercept is =. Plotting these two intercept points and connect the dots to graph the line:. Rewrite the equation is slope-intercept form. x + = 6 x x = x + 6 x 6 = + = x + The x-intercept occurs when = 0, and the -intercept occurs when x = 0. When = 0 we have: 0 = x + = x x = = x = x So the x-intercept is x =. When x = 0 we have: = 0 + = 0+ = So the -intercept is =. Plotting these two intercept points and connect the dots to graph the line: Page of
Chapter Section : Equations of Lines 7. Rewrite the equation in slope-intercept form: x + = 6 When = 0 we have: 0 = x + = x x = = x = x So, the x-intercept is x =. x x = x + 6 x 6 = + = x + When x = 0 we have: = 0 + = 0+ = So, the -intercept is =. Plotting these two intercept points and connect the dots to graph the line: 9. This line goes through the points (0,) and (,0). So it has slope 0 m = = =. 0 Since the line has -intercept, the line s equation is = x +. Page of