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Five-Minute Check (over Lesson 3–5) CCSS Then/Now New Vocabulary Key Concept: Distance Between a Point and a Line Postulate 3.6: Perpendicular Postulate Example 1: Real-World Example: Construct Distance From Point to a Line Example 2: Distance from a Point to a Line on Coordinate Plane Key Concept: Distance Between Parallel Lines Theorem 3.9: Two Line Equidistant from a Third Example 3: Distance Between Parallel Lines Lesson Menu

Mathematical Practices 2 Reason abstractly and quantitatively. Content Standards G.CO.12 Make formal geometric constructions with a variety of tools and methods (compass and straightedge, string, reflective devices, paper folding, dynamic geometric software, etc.). G.MG.3 Apply geometric methods to solve problems (e.g., designing an object or structure to satisfy physical constraints or minimize cost; working with typographic grid systems based on ratios). Mathematical Practices 2 Reason abstractly and quantitatively. 4 Model with mathematics. CCSS

Find the distance between a point and a line. You proved that two lines are parallel using angle relationships. Find the distance between a point and a line. Find the distance between parallel lines. Then/Now

equidistant Vocabulary

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Construct Distance From Point to a Line CONSTRUCTION A certain roof truss is designed so that the center post extends from the peak of the roof (point A) to the main beam. Construct and name the segment whose length represents the shortest length of wood that will be needed to connect the peak of the roof to the main beam. The distance from a line to a point not on the line is the length of the segment perpendicular to the line from the point. Locate points R and S on the main beam equidistant from point A. Example 1

Construct Distance From Point to a Line Locate a second point not on the beam equidistant from R and S. Construct AB so that AB is perpendicular to the beam. Answer: The measure of AB represents the shortest length of wood needed to connect the peak of the roof to the main beam. ___ Example 1

KITES Which segment represents the shortest distance from point A to DB? A. AD B. AB C. CX D. AX Example 1

Step 1 Find the slope of line s. Distance from a Point to a Line on Coordinate Plane COORDINATE GEOMETRY Line s contains points at (0, 0) and (–5, 5). Find the distance between line s and point V(1, 5). (–5, 5) (0, 0) V(1, 5) Step 1 Find the slope of line s. Begin by finding the slope of the line through points (0, 0) and (–5, 5). Example 2

The equation of line s is y = –x. Distance from a Point to a Line on Coordinate Plane Then write the equation of this line by using the point (0, 0) on the line. Slope-intercept form m = –1, (x1, y1) = (0, 0) Simplify. The equation of line s is y = –x. Example 2

Subtract 1 from each side. Distance from a Point to a Line on Coordinate Plane Step 2 Write an equation of the line t perpendicular to line s through V(1, 5). Since the slope of line s is –1, the slope of line t is 1. Write the equation for line t through V(1, 5) with a slope of 1. Slope-intercept form m = 1, (x1, y1) = (1, 5) Simplify. Subtract 1 from each side. The equation of line t is y = x + 4. Example 2

2y = 4 Add the two equations. y = 2 Divide each side by 2. Distance from a Point to a Line on Coordinate Plane Step 3 Solve the system of equations to determine the point of intersection. line s: y = –x line t: (+) y = x + 4 2y = 4 Add the two equations. y = 2 Divide each side by 2. Solve for x. 2 = –x Substitute 2 for y in the first equation. –2 = x Divide each side by –1. The point of intersection is (–2, 2). Let this point be Z. Example 2

Distance from a Point to a Line on Coordinate Plane Step 4 Use the Distance Formula to determine the distance between Z(–2, 2) and V(1, 5). Distance formula Substitution Simplify. Answer: The distance between the point and the line is or about 4.24 units. Example 2

COORDINATE GEOMETRY Line n contains points (2, 4) and (–4, –2) COORDINATE GEOMETRY Line n contains points (2, 4) and (–4, –2). Find the distance between line n and point B(3, 1). B(3, 1) (2, 4) (–4, 2) A. B. C. D. Example 2

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Distance Between Parallel Lines Find the distance between the parallel lines a and b whose equations are y = 2x + 3 and y = 2x – 1, respectively. You will need to solve a system of equations to find the endpoints of a segment that is perpendicular to both a and b. From their equations, we know that the slope of line a and line b is 2. Sketch line p through the y-intercept of line b, (0, –1), perpendicular to lines a and b. a b p Example 3

Subtract 1 from each side. Distance Between Parallel Lines Step 1 Write an equation for line p. The slope of p is the opposite reciprocal of Use the y-intercept of line b, (0, –1), as one of the endpoints of the perpendicular segment. Point-slope form Simplify. Subtract 1 from each side. Example 3

Substitute 2x + 3 for y in the second equation. Distance Between Parallel Lines Step 2 Use a system of equations to determine the point of intersection of the lines a and p. Substitute 2x + 3 for y in the second equation. Group like terms on each side. Example 3

Substitute for x in the equation for p. Distance Between Parallel Lines Simplify on each side. Multiply each side by . Substitute for x in the equation for p. Example 3

The point of intersection is or (–1.6, –0.2). Distance Between Parallel Lines Simplify. The point of intersection is or (–1.6, –0.2). Example 3

Answer: The distance between the lines is about 1.79 units. Distance Between Parallel Lines Step 3 Use the Distance Formula to determine the distance between (0, –1) and (–1.6, –0.2). Distance Formula x2 = –1.6, x1 = 0, y2 = –0.2, y1 = –1 Answer: The distance between the lines is about 1.79 units. Example 3

Find the distance between the parallel lines a and b whose equations are and , respectively. A. 2.13 units B. 3.16 units C. 2.85 units D. 3 units Example 3

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