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COORDINATE GEOMETRY Find the distance from P to Line FRQWDLQVSRLQWV ±2, 1) and (4, 1). Point P has coordinates (5, 7). 6
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3-6 Perpendiculars and Distance COORDINATE GEOMETRY Find the distance from P to 17. Line contains points (–2, 1) and (4, 1). Point P has coordinates (5, 7). SOLUTION: Use the slope formula to find the slope of the line
.Let (x 1, y 1) = (–2, 1)and (x2, y 2) = (4, 1).
Use the slope and any one of the points to write the equation of the line. Let (x 1, y 1) = (–2, 1).
The slope of an equation perpendicular to will be undefined, and hence the line will be a vertical line. The equation of a vertical line through (5, 7) is x = 5. The point of intersection of the two lines is (5, 1).
Use the Distance Formula to find the distance between the points (5, 1) and (5, 7). Let (x 1, y 1) = (5, 1) and (x2, y 2) = (5, 7).
Therefore, the distance between the line and the point is 6 units. 19. Line
contains points (1, 5) and (4, –4). Point P has coordinates (–1, 1).
SOLUTION: Use the slope formula to find the slope of the line
. Let (x 1, y 1) = (1, 5) and (x2, y 2) = (4, –4),
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Use the slope and any one of the points to write the equation of the line. Let (x 1, y 1) = (4, –4).
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19. Line
contains points (1, 5) and (4, –4). Point P has coordinates (–1, 1).
3-6 Perpendiculars SOLUTION: and Distance Use the slope formula to find the slope of the line
. Let (x 1, y 1) = (1, 5) and (x2, y 2) = (4, –4),
Use the slope and any one of the points to write the equation of the line. Let (x 1, y 1) = (4, –4).
The slope of an equation perpendicular to l will be
So, write the equation of a line perpendicular to l and that
passes through (–1, 1).
Solve the system of equations to determine the point of intersection.
The left sides of the equations are the same. So, equate the right sides and solve for x.
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Solve the systemand of equations 3-6 Perpendiculars Distanceto determine the point of intersection.
The left sides of the equations are the same. So, equate the right sides and solve for x.
Use the value of x to find the value of y.
So, the point of intersection is (2, 2).
Use the Distance Formula to find the distance between the points (2, 2) and (–1, 1).
Therefore, the distance between the line and the point is Find the distance between each pair of parallel lines with the given equations. 23. SOLUTION: To find the distance between the parallel lines, we need to find the length of the perpendicular segment between the two parallel lines. Pick a point on one of the equation, and write the equation of a line perpendicular through that point. Then use this perpendicular line and other equation to find the point of intersection. Find the distance between the two point using the distance formula.
Step 1: Find the equation of the line perpendicular to each of the lines.
The slope of a line perpendicular to both the lines will be
Consider the y-intercept of any of the two lines and
write the equation of the perpendicular line through it. The y-intercept of the line y = 5x + 4 is (0, 4). So, the equation eSolutions Manual - Powered by Cognero
of a line with slope
and a y-intercept of 4 is
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Step 1: Find the equation of the line perpendicular to each of the lines.
3-6 Perpendiculars and Distance The slope of a line perpendicular to both the lines will be
Consider the y-intercept of any of the two lines and
write the equation of the perpendicular line through it. The y-intercept of the line y = 5x + 4 is (0, 4). So, the equation of a line with slope
and a y-intercept of 4 is
Step 2: Find the intersections of the perpendicular line and each of the other lines. To find the point of intersection of the perpendicular and the second line, solve the two equations. The left sides of the equations are the same. So, equate the right sides and solve for x.
Use the value of x to find the value of y.
So, the point of intersection is (5, 3).
Step 3: Find the length of the perpendicular between points Use the Distance Formula to find the distance between the points (5, 3) and (0, 4). Let (x 1, y 1) = (5, 3) and (x2, y 2) = (0, 4).
Therefore, the distance between the two lines is
27. SOLUTION: To find the distance between the parallel lines, we need to find the length of the perpendicular segment between the eSolutions Manual - Powered by Cognero two parallel lines. Pick a point on one of the equation, and write the equation of a line perpendicular through that Page 4 point. Then use this perpendicular line and other equation to find the point of intersection. Find the distance between the two point using the distance formula.
27. 3-6 Perpendiculars and Distance SOLUTION: To find the distance between the parallel lines, we need to find the length of the perpendicular segment between the two parallel lines. Pick a point on one of the equation, and write the equation of a line perpendicular through that point. Then use this perpendicular line and other equation to find the point of intersection. Find the distance between the two point using the distance formula.
Step 1: Find the equation of the line perpendicular to each of the lines. First, write the second equation also in the slope-intercept form.
The slope of a line perpendicular to both the lines will be –4. Consider the y-intercept of any of the two lines and write the equation of the perpendicular line through it. The y-intercept of the line
is (0, 2). So, the
equation of a line with slope –4 and a y-intercept of 2 is
Step 2: Find the intersections of the perpendicular line and each of the other lines. To find the point of intersection of the perpendicular and the second line, solve the two equations. The left sides of the equations are the same. So, equate the right sides and solve for x.
Use the value of x to find the value of y.
So, the point of intersection is (4, –14).
Step 3: Find the length of the perpendicular between points. Use the Distance Formula to find the distance between the points (4, –14) and (0, 2). Let (x 1, y 1) = (4, –14) and (x2, y2) = (0, 2). eSolutions Manual - Powered by Cognero
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So, the point of intersection is (4, –14).
Step 3: Find the length of the perpendicular between points. 3-6 Perpendiculars Distance Use the Distanceand Formula to find the distance between the points (4, –14) and (0, 2). Let (x 1, y 1) = (4, –14) and (x2, y2) = (0, 2).
Therefore, the distance between the two lines is CONSTRUCTION Line contains points at (–4, 3) and (2, –3). Point P at (–2, 1) is on line . Complete the following construction. Step 1 Graph line and point P, and put the compass at point P. Using the same compass setting, draw arcs to the left and right of P. Label these points A and B.
Step 2 Open the compass to a setting greater than AP . Put the compass at point A and draw an arc above line
Step 3 Using the same compass setting, put the compass at point B and draw an arc above line point of intersection Q. Then draw .
36. What is the relationship between line
and
.
. Label the
? Verify your conjecture using the slopes of the two lines.
SOLUTION: Sample answer: The lines are perpendicular; the slope of negative reciprocals, the lines are perpendicular.
is –1 and the slope of
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37. Repeat the construction above using a different line and point on that line.
is 1. Since the slopes are Page 6
3-6 Perpendiculars and Distance Therefore, the distance between the two lines is CONSTRUCTION Line contains points at (–4, 3) and (2, –3). Point P at (–2, 1) is on line . Complete the following construction. Step 1 Graph line and point P, and put the compass at point P. Using the same compass setting, draw arcs to the left and right of P. Label these points A and B.
Step 2 Open the compass to a setting greater than AP . Put the compass at point A and draw an arc above line
Step 3 Using the same compass setting, put the compass at point B and draw an arc above line point of intersection Q. Then draw .
36. What is the relationship between line
and
.
. Label the
? Verify your conjecture using the slopes of the two lines.
SOLUTION: Sample answer: The lines are perpendicular; the slope of negative reciprocals, the lines are perpendicular.
is –1 and the slope of
is 1. Since the slopes are
37. Repeat the construction above using a different line and point on that line. SOLUTION: Sample answer: Step 1: Graph line through points (–2, –4), (2, –2), and (4, –1) with Point P at (–2, –4).Put the compass at point P. Using the same compass setting, draw arcs to the left and right of P. Label these points A and B. eSolutions Manual - Powered by Cognero
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SOLUTION: Sample answer: The lines are perpendicular; the slope of 3-6 Perpendiculars and the Distance negative reciprocals, lines are perpendicular.
is –1 and the slope of
is 1. Since the slopes are
37. Repeat the construction above using a different line and point on that line. SOLUTION: Sample answer: Step 1: Graph line through points (–2, –4), (2, –2), and (4, –1) with Point P at (–2, –4).Put the compass at point P. Using the same compass setting, draw arcs to the left and right of P. Label these points A and B.
Step 2: Open the compass to a setting greater than AP. Put the compass at point A and draw an arc above the line.
Step 3: Using the same compass setting, put the compass at point B and draw an arc above the line. Label the point of intersection Q. Then draw a line through Q and P.
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