Coordinate Geometry: Chapter 1 PDF

CHAPTER 1 COORDINATE GEOMETRY CHAPTER 1 Coordinate Geometry Coordinate Geometry is one of the most fascinating mathematical concepts. Coordinate geometry, also known as analytic geometry, describes the relationship between geometry and algebra using graphs that include curves and l...

CHAPTER 1 COORDINATE GEOMETRY CHAPTER 1 Coordinate Geometry Coordinate Geometry is one of the most fascinating mathematical concepts. Coordinate geometry, also known as analytic geometry, describes the relationship between geometry and algebra using graphs that include curves and lines. It teaches geometric aspects of Algebra and allows students to solve geometric problems. It is a branch of geometry in which the position of points on a plane is represented by an ordered pair of numbers. 1.1 Cartesian coordinate system The cartesian coordinate system is made up of a number line called the cartesian plane that is divided into four quadrants by two axes that are perpendicular to one another and are designated as the x-axis for the horizontal line and the y-axis for the vertical line. A location on the coordinate plane that known as origin is where the x- and y-axes cross. Figure 1.1(b) shows the four quadrants along with the corresponding values. A pair of numbers (x, y) that describe a point's location on a plane are known as its coordinates. The example of coordinates in the Cartesian plane are visualised as (3, 4) in Figure 1.1(a). The distance between the two points and the interval's midpoint that connects the points can be calculated if the coordinates are known. 1  2nd Quadrant 1st Quadrant x (–ve) & y (+ve) x (+ve) & y (+ve) 3rd Quadrant 4th Quadrant x (–ve) & y (–ve) x (+ve) & y (–ve) (a) (b) Figure 1.1: Four quadrants in Cartesian plane 1.2 Coordinate geometry formulas. The study of geometry using coordinate points is known as coordinate geometry or analytic geometry. Using coordinate geometry, it is possible to calculate the distance between two points, the midpoint of a line, the m:n ratio of a line, the area of a triangle in the Cartesian plane, etc., where the formulae are listed below with the aid of a diagram that illustrated in Figure 1.2. Figure 1.2: Coordinate geometry with three points A, B and C 2 Let’s consider the coordinate geometry given in Figure 1.2. 1. Distance formula: To calculate distance between two points. Distance from point A to B, d = √(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2 2. Section formula: To find a point which divides a Line into ℓ: 𝑘 ratio. ℓ𝑥1 +𝑘𝑥2 ℓ𝑦1` +𝑘𝑦2 If ℓ ≠ 𝑘 then, Point C = ( , ) ℓ+𝑘 𝑙+𝑘 𝑥1 +𝑥2 𝑦1` +𝑦2 If ℓ = 𝑘 then point C is a Mid Point = ( 2 , 2 ) 3. Formula of Area: To find an area when given four coordinates in the Cartesian plane which are A (𝑥1 , 𝑦1 ) , B (𝑥2 , 𝑦21 ) , C (𝑥3 , 𝑦3 ) , and D(𝑥4 , 𝑦4 ). 1 𝑥1 𝑥2 𝑥3 𝑥1 Triangle ABC = 2 |𝑦 𝑦2 𝑦3 𝑦1 | 1 1 = 2 [(𝑥1 𝑦2 + 𝑥2 𝑦3 + 𝑥3 𝑦1 ) − (𝑥1 𝑦3 + 𝑥3 𝑦2 + 𝑥2 𝑦1 )] 1 𝑥1 𝑥2 𝑥3 𝑥4 𝑥1 Rectangle ABCD = |𝑦 𝑦2 𝑦3 𝑦4 𝑦1 | 2 1 1 = 2 [(𝑥1 𝑦2 + 𝑥2 𝑦3 + 𝑥3 𝑦4 + 𝑥4 𝑦1 ) − (𝑥2 𝑦1 + 𝑥3 𝑦2 + 𝑥4 𝑦3 + 𝑥1 𝑦4 )] 4. Slope formula: To find a slope of a line joining two points A and B. y 2 − y1 m= = tan  x 2 − x1 3 Example 1.1 Find the lengths of the straight lines joining each of the following pairs of points. a. (4, –1) and (–3, –4) c. (a, b) and (b, a) b. (√6, √3) and (3√6, −√3) d. (5, 1) and (9, -3) Solution: a. √(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2 = √(−3 − 4)2 + (−4 + 1)2 = √58 2 2 b. √(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2 = √(3√6 − √6) + (−√3 − √3) = √36 = 6 c. √(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2 = √(𝑏 − 𝑎)2 + (𝑎 − 𝑏)2 = √2𝑎2 − 4𝑎𝑏 + 𝑏 2 d. √(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2 = √(9 − 5)2 + (−3 − 1)2 = 4√2 Example 1.2 The points A, B and C have coordinates (2, 1), (7, 3) and (5, k) respectively. If AB and BC are equal length, find the possible values of k. Solution: 𝐴𝐵 = 𝐵𝐶 √(7 − 2)2 + (3 − 1)2 = √(5 − 7)2 + (𝑘 − 3)2 √29 = √4 + 𝑘 2 − 6𝑘 + 9 𝑘 2 − 6𝑘 − 16 = 0 (𝑘 − 8)(𝑘 + 2) = 0 ∴ 𝑘 = 8 𝑎𝑛𝑑 𝑘 = −2 4 Example 1.3 Given both distances from origin to points (p, 2p) and (p - 3, 2p + 1) are same. Determine the possible values of p. Solution: √(𝑝 − 0)2 + (2𝑝 − 0)2 = √((𝑝 − 3) − 0)2 + ((2𝑝 + 1) − 0)2 𝑝2 + 4𝑝2 = (𝑝2 − 6𝑝 + 9) + (4𝑝2 + 4𝑝 + 1) 0 = −2𝑝 + 10 2𝑝 = 10 𝑝=5 EXERCISE 1.1 Calculate the lengths of the straight line between point 𝐴 (ℎ, 5ℎ) and 𝐵 (3ℎ, −2ℎ). If 𝐴𝐵 = 5 53 , determine the value of ℎ. Solution: 5 EXERCISE 1.2 The three points 𝑂, 𝐴, and 𝐵 have coordinates (0, 0), (𝑎, 7) and (8, 1) respectively. If 𝑂𝐴 and 𝑂𝐵 are equal length, calculate the two possible values of 𝑎. Solution: EXERCISE 1.3 The three points 𝐺 (4, 0), 𝐻 (ℎ, 6) and 𝐼 (7, 1) are given, as such the length of 𝐺𝐻 is twice 𝐺𝐼. Calculate the two possible values of ℎ. Solution: 6 Example 1.4 Determine the midpoint’s coordinates of the straight lines joining each of the following pairs of points: a. (4, – 1) and (– 3, – 4) c. (𝑎, 𝑏) and (𝑏, 𝑎) b. (√6, √3) and (3√6, −√3) d. (5, 1) and (9, −3) Solution: 𝑥1 +𝑥2 𝑦1 +𝑦2 4−3 −1−4 1 5 a. midpoint = ( 2 , 2 )=( 2 , 2 ) = ( ,− ) 2 2 𝑥1 +𝑥2 𝑦1 +𝑦2 √6+3√6 √3−√3 4√6 0 b. midpoint = ( 2 , 2 )=( 2 , 2 )=( 2 , ) = (2√6, 0) 2 𝑥1 +𝑥2 𝑦1 +𝑦2 𝑎+𝑏 𝑏+𝑎 c. midpoint = ( 2 , 2 )=( 2 , 2 ) 𝑥1 +𝑥2 𝑦1 +𝑦2 5+9 1−3 14 −2 d. midpoint = ( , )=( , )=( , ) = (7, −1) 2 2 2 2 2 2 Example 1.5 The points 𝑃 (4, −3), 𝑄 (−3, 4), 𝑅 (−2, 7) and 𝑆 are the vertices of a parallelogram 𝑃𝑄𝑅𝑆. Determine: a. The midpoint’s coordinates of the diagonal 𝑃𝑅. b. The coordinates of 𝑆. Solution: 𝑥1 +𝑥2 𝑦1 +𝑦2 4−2 −3+7 2 4 a. ( 2 , 2 ) =( 2 , 2 ) = (2 , 2) = (1,2) −3+𝑥 4+𝑦 b. midpoint 𝑄𝑆 = midpoint 𝑃𝑅, therefore ( 2 , 2 ) = (1,2). Compare −3+𝑥 4+𝑦 each coordinate: = 1 → 𝑥 = 5, and = 2 → 𝑦 = 0. ∴ 𝑆 = (5,0) 2 2 7 Example 1.6 𝑃(3, 4) is a point on a straight line that connect two points which are 𝐴 (−1, 2) and 𝐵 (9, 7). Determine the ratio of 𝐴𝑃: 𝑃𝐵. Solution: Let's say 𝑃 divide 𝐴𝐵 with ratio 𝑚 ∶ 𝑛. 𝑛 × (−1) + 𝑚 ×9 𝑛 ×2 + 𝑚 ×7 −𝑛+9𝑚 2𝑛+7𝑚 Then, 𝑃 = ( , ) =( , ) 𝑚+𝑛 𝑚+𝑛 𝑚+𝑛 𝑚+𝑛 Given 𝑃 is (3, 4). −𝑛+9𝑚 2𝑛+7𝑚 Then, =3 and =4 𝑚+𝑛 𝑚+𝑛 −𝑛 + 9𝑚 = 3𝑚 + 3𝑛 2𝑛 + 7𝑚 = 4𝑚 + 4𝑛 6𝑚 = 4𝑛 3𝑚 = 2𝑛 𝑚 2 𝑚 2 𝑛 =3 𝑛 =3 EXERCISE 1.4 Find the coordinates of the point 𝑃 which divides the line joining 𝐴 (3, 2) and 𝐵 (5, 1) in the ratio 2: 3. Solution: 8 EXERCISE 1.5 Given 𝐴 (−2, 0), 𝐵 (1, 2) and 𝐶 (7, 6). Determine the ratio that divides the line by the point. a. 𝐵 on a line 𝐴𝐶 b. 𝐴 on a line 𝐵𝐶 Solution: Example 1.7 Determine the area of triangle that has vertices at 𝐴 (−2, 1), 𝐵 (0, 5) and 𝐶 (2, −2). Solution: 1 −2 0 2 −2 Area of triangle 𝐴𝐵𝐶 = | 2 1 | 5 −2 1 1 =2 |[(−2(5)) + (0)(−2) + (2)(1)] − [(−2)(−2) + (2)(5) + (0)(1)]| 1 = |(−10 + 0 + 2) − (4 + 10 + 0)| 2 = 11 unit2 9 Example 1.8 Points A, B, C and D have a coordinate (-3, 8), (p, 3), (9, 2) and (4, q) respectively. Given ABCD is a parallel square. Determine the area of ABCD. Solution: −3+9 8+2 S1: Determine the middle point for AC = ( 2 , 2 ) = (3, 5) 𝑝+4 3+𝑞 S2: Determine the middle point for BD = ( 2 , 2 ) 𝑝+4 3+𝑞 S3: ( 2 , 2 ) = (3, 5) 𝑝+4 3+𝑞 Then, =3 and =5 2 2 p+4=6 3 + q = 10 p=2 q=7 S4: Area for parallel square A (-3, 8), B (2, 3), C (9, 2) dan D (4, 7) Area ABCD = 2 x area of triangle BCD 1 2 9 4 2 = 2x | 2 3 | 2 7 3 1 = 2 × (30) 2 = 30 unit2 EXERCISE 1.6 Prove that points 𝐴 (−2, 0), 𝐵 (1, 2) and 𝐶 (4, 4) is on the same lines. 10 Solution: EXERCISE 1.7 Given the points 𝑃 (𝑎, 𝑏), 𝑄 (−3, 2) and 𝑅 (−4, −4). If the area of 35 triangle 𝑃𝑄𝑅 is 2 , prove that 6𝑎 – 𝑏 – 15 = 0. Solution: 11 EXERCISE 1.8 Determine the area of rectangle that has vertices at A (3, -1), B (2, 2), C (-4, 3) and D (-1, -4). Solution: Example 1.9 Find the slope of the lines joining the following pairs of points: a. (1, 3) and (4, 6) b. (-1, -5) and (2, -3) c. (t, 0) and (0, t) Solution: 𝑦 −𝑦 6−3 a. 𝑚 = 𝑥2 −𝑥1 = 4−1 = 1 2 1 −3−(−5) 2 b. 𝑚 = 2−(−1) =3 𝑡−0 𝑡 c. 𝑚 = 0−𝑡 = −𝑡 = −1 12 EXERCISE 1.9 Determine the slope of the line through origin with intersection points between the curve 𝑦 = 2𝑥 2 and line 𝑦 = 𝑥 + 3. Solution EXERCISE 1.10 The points 𝐴 (5, 2), 𝐵 (1, 0), 𝐶 (𝑐, 5) and 𝐷 (𝑑, −2) all lie on the same straight line. Find the values of 𝑐 and 𝑑. Solution: 13 1.3 Slope and intercept of a line. 1.3.1 Slope of a line A slope of a line is a numerical measurement of the line “steepness”. It is typically represented by the letter 𝑚, and represent a change in y along the line for a unit change in x. A line's slope, often known as its gradient, is a numerical measure of how "steep" it is. In section 1.2, the formula to determine the slope of a line connecting two locations A and B is provided. It is irrelevant which point is selected as A or B. The formula will yield the right slope if they are both somewhere along the line. The slope can also be calculated by comprehending the idea and working it out analytically, as opposed to just entering numbers into the formula. See the diagram on the side where a line is shown as being defined by the two provided points A and B. The slope will be positive due to the line's upward and rightward slope. Below are the steps in calculating the slope of above example: 1. Determine the run (𝑑𝑥), the horizontal distance between the left and right points. 𝐵 is located at (15,5), hence the first number in its 𝑥- coordinate is 15. A has an 𝑥-coordinate of 30. The run, (𝑑𝑥) is therefore 15. 2. Determine the rise (𝑑𝑦), the amount of the line rises or lowers as you move to the right. Based on the above figure, y-coordinate for 𝐵 is 5 14  since it is a second number for (15,5), as such y-coordinate of 𝐴 is 25. Therefore, the rise is +20. The rise is positive because the line goes up as you go to the right. It would have been negative otherwise. +20 3. Dividing the rise (𝑑𝑦) by the run (𝑑𝑥): Slope = 15 = +1.33 A way to remember this method is "rise over run", where the "rise" is the up and down difference between the points, and the "run" is the horizontal run between them. The slope of the line can also be expressed as an angle, usually in degrees or radians. Equation 1.1 shows how to convert from slope m to slope angle and vice versa. 𝜃 = tan−1(𝑚) 𝑚 = tan (𝜃) Equation 1.1: Slope in angle The slope of a line can positive, negative, zero or undefined as shown in Figure 1.3. Positive Negative Zero Undefined Figure 1.3: Illustration of positive, negative, zero and undefined slope. 15 1.3.2 Intercept of a line The intercept of a line is the y-coordinate of the point that crosses the 𝑦- axis. The intercept of a line is the point at which it crosses either the 𝑥 or 𝑦 axis. If the cross axes are not specified, the 𝑦-axis is assumed. It is usually denoted by the letter 𝑏. Nevertheless, the exactly vertical line will always cross the 𝑦-axis at any point on the coordinate plane. If the value of slope m and one point (𝑥, 𝑦) in a line are given, then b can be calculated using formula: 𝑏 = 𝑦 − 𝑚𝑥 If two points in a line (𝑥1 , 𝑦1 ) and (𝑥2 , 𝑦2 ) are given, then the slope, m need to be determined using the method describe in section 1.2 before the intercept, 𝑏 can be calculated using the above formula. 1.4 Equation of a line in Cartesian plane 1.4.1 Equation of horizontal line On the coordinate plane, a horizontal line is a straight line where every point has the same y-coordinate. It is one that parallel with the x-axis of the coordinate plane. The y-coordinate of each point along the line will be the same. The slope of a horizontal line is zero. Moving from the left to the right along the line, does not give the effect of rise or fall. Therefore, two points on the horizontal line will have the same y-coordinate, and the slope is zero. The equation of a horizontal line is 𝑦 = 𝑏, where for any x coordinates the line will cross the y-axis at 𝑏. Notice that the equation is independent of x. Any point on the horizontal line satisfies the equation. 16 Example The two points A, B on the line are at (7,7) and (39,7). The second coordinate in each pair is the y-coordinate which are 7, and 7. Since they are equal, the line is horizontal. Since the line passes the y-axis at 7, the equation of the line is 𝑦 = 7 which can be read as "for all values of x, y is 7". 1.4.2 Equation of vertical line A vertical line is a line on the coordinate plane where all points on the line have the same x-coordinate. It is a line that is parallel to the y-axis of coordinate plane and runs straight up and down. The identical x-coordinate will be shared by every point along the line. A vertical line has no slope or in other word, the slope of a vertical line is undefined. The equation of a vertical line is 𝑥 = 𝑎 where for any 𝑦 coordinates the line will cross the 𝑥-axis at 𝑎. Notice that the equation is independent of y. Any point on the vertical line satisfies the equation. 17 Example The points A and B on the line are at (-15,3) and (-15,20). The first coordinate in each pair is the x-coordinate which are -15, and -15. Since they are equal, the line is vertical. Since the line crosses the x-axis at -15, the equation of the line is 𝑥 = 15 , which can be read as "for all values of y, x is -15". Example 1.10 Determine the equation of the lines below: a. Vertical line pass through the point 𝐴 (2, −1) Answer: ____________________ b. Horizontal line with distance -3 from 𝑥-axis Answer: ____________________ 1.4.3 Equation of a line (Slope-intercept form) A straight line on the coordinate plane can be described by the equation 𝑦 = 𝑚𝑥 + 𝑐 where: 𝑚 = the slope of the line 18  𝑏 = the intercept (where the line crosses the y-axis) 𝑥, 𝑦 = the coordinates of any point on the line. Recall that the slope 𝑚 is the "steepness" of the line and 𝑏 is the intercept which is the point where the line crosses the 𝑦-axis. Example 1.11 Determine a slope of the following straight lines, and in each case find the coordinates of a point where the line cuts the 𝑦-axis. a. y = 3x + 4 b. 2y = 3x – 4 c. x + 4y = 2 Solution a. y = 3x + 4: 𝑚 = 3, 𝑏 = 4, then slope, 𝑚 is 3 and intercept is (0, 4). 3 3 b. 𝑦 = 2 𝑥 − 2: 𝑚 = 2, 𝑏 = −2, 3 then slope, 𝑚 is 2 and intercept is (0, −2). 1 1 1 1 c. 4𝑦 = −𝑥 + 2 → 𝑦 = − 4 𝑥 + 2: 𝑚 = − 4, 𝑏 = 2 1 1 then slope, 𝑚 is − and intercept is (0, ) 4 2 19 Example 1.12 Determine the point at which the line 3𝑦 – 12 = 4𝑥 cuts the 𝑦-axis, the 𝑥-axis and hence sketch the line 3𝑦 – 12 = 4𝑥. Solution: 4 3𝑦 − 12 = 4𝑥 ----> 𝑦 = 3 𝑥 + 4 when 𝑥 = 0, 𝑦 = 4, so intercept at y-axis is (0,4) when 𝑦 = 0, 𝑥 = −3, so intercept at x-axis is (−3,0) 1.4.4 Equation of the line (Point-slope form) Point slope form is used to represent a straight line using its slope and a point on the line. That means, the equation of a line whose slope is 'm' and which passes through a point (𝑥1 , 𝑦1 ) is found using the point slope form. The equation of the point slope form is 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ) where: (𝑥, 𝑦) is a random point on the line. (𝑥1 , 𝑦1 ) is a fixed point on the line. 𝑚 is the slope of the line. 20 To solve point slope form for a given straight line for finding the equation of the given line, we can follow the steps given below, Step 1: Note down the slope, 'm' of the straight line, and the coordinates (𝑥1 , 𝑦1 ) of the given point that lies on the line. Step 2: Substitute the given values in the point slope formula: 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ) Step 3: Simplify to obtain the equation of the line in standard form. EXERCISE 1.11 Determine the equation of the line that pass through the intersection point between the lines 𝑥 + 2𝑦 = 8 and 2𝑦 – 𝑥 = 8 with the slope 6. Solution 21 1.4.5 Equation of a line through 2 points Given two points 𝐴(𝑥1 , 𝑦1 ) and 𝐵(𝑥2 , 𝑦2 ), then the equation of line can be determined using formula 𝑦 − 𝑦1 𝑦2 − 𝑦1 = 𝑥 − 𝑥1 𝑥2 − 𝑥1 where: (𝑥, 𝑦) is a random point on the line (𝑥1 , 𝑦1 ) and (𝑥2 , 𝑦2 ) are two fixed points on the line. Below are the steps in finding the equation of straight line given two points on the line: Step 1: Note down the coordinates (𝑥1 , 𝑦1 ) and (𝑥2 , 𝑦2 ) of the given points. 𝑦−𝑦1 𝑦2 −𝑦1 Step 2: Substitute the given coordinate values in formula: = 𝑥−𝑥1 𝑥2 −𝑥1 Step 3: Simplify to obtain the equation of the line in standard form. EXERCISE 1.12 Find the equation of the straight line that pass through the points. a. (2, 1) and (-2, -7) c. (1, -2) and (-4, 1) b. (6, -2) and (12, 1) d. (0, 7) and (7, 0) 22 Solution 23 1.5 Rules of a line in Cartesian plane 1.5.1 Parallel and perpendicular lines If two lines do not intersect, they are said to be parallel. They both have the same steepness, hence they both have the same slopes. The y-intercept is the only distinction between the two lines. If one of the lines is shifted vertically toward the y-intercept of the other, they would become the same line. The identification of parallel lines can be done by comparing the slopes of given equations. The lines are parallel if the slopes are the same and the y-intercepts are different. The lines are not parallel if the slopes differ. a) Parallel lines b) Perpendicular lines Figure 1.4: Illustration of parallel and perpendicular lines Perpendicular lines, unlike parallel lines, do intersect. Their intersection creates a 90-degree angle. The slopes of perpendicular lines are not the 24 same. Perpendicular line slopes differ from one another in a specific way. The slope of one line is equal to the inverse of the slope of the other line. A number's product with its reciprocal is 1. If 𝑚1 and 𝑚2 are negative reciprocals of each other, they can be multiplied to yield 1. Generally, given two equations of the lines 𝑓(𝑥) = 𝑚1 𝑥 + 𝑏1 and 𝑔(𝑥) = 𝑚2 𝑥 + 𝑏2. 1. Lines 𝑓(𝑥) and 𝑔(𝑥) can be said as parallel lines if 𝑚1 = 𝑚2. 2. Lines 𝑓(𝑥) and 𝑔(𝑥) can be said as parallel lines if 𝑚1 𝑚2 = −1 1 where 𝑚2 = −. 𝑚1 Example 1.13 The line 𝑦 = 𝑎𝑥 + 𝑏 is parallel to the line 𝑦 = 2𝑥– 6 and passes through the point (−1, 7), find the value of 𝑎 and 𝑏. Solution: Parallel lines therefore 𝑎 = 2. Then substitute the value a, x and y in 𝑦 = 𝑎𝑥 + 𝑏, such that 𝑏 = 𝑦 − 𝑎𝑥 = 7 − 2(−1) = 9 Example 1.14 Determine the equation of a straight line that parallel with line 3𝑥 + 2𝑦 + 8 = 0 and pass through the intersection point of lines 3𝑥 − 2𝑦 + 1 = 0 and 𝑥 + 𝑦 + 2 = 0. 25 Solution: 3 S1: From 3𝑥 + 2𝑦 + 8 = 0 ---------> 𝑦 = − 2 𝑥 − 4, 3 therefore, the slope is − 2. S2: Determine the intersection point. 3x − 2y + 1 = 0 -----------(1) x + y + 2 = 0---------------(2) (2) x 2: 2x + 2y + 4 = 0------------(3) (1) + (3): 5𝑥 = −5 𝑥 = −1 substitute 𝑥 = −1 in (1): 𝑦 = −1 3 Therefore, the intersection point is (-1, -1) and 𝑚 = − 2 EXERCISE 1.13 The line 𝑦 = 𝑐𝑥 + 𝑑 is perpendicular to the line 𝑦 + 2𝑥 = 4 and cuts the line 𝑦 + 𝑥 + 3 = 0 and the 𝑦-axis, find the value of 𝑐 and 𝑑. Solution: 26 EXERCISE 1.14 Determine the equation of the straight line that pass through the point a. (4, 3) and parallel with line 3𝑥 – 4𝑦 + 6 = 0 b. (4, -1) and perpendicular with line 𝑥 + 7𝑦 – 4 = 0 Solution: Example 1.15 Find the equation of the line that pass through the intersection point between 3𝑦– 𝑥 = 2 and 2𝑦– 5𝑥 = 1 and perpendicular with line 𝑦– 2𝑥 = 7. Solution: S2: 1 1 9 𝑚 = − and pass through ( , ) 2 13 13 S1: 3𝑦– 𝑥 = 2--------(1) 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ) 2𝑦– 5𝑥 = 1------(2) 9 1 1 (1)x5: 15𝑦– 5𝑥 = 10----(3) 𝑦− = − (𝑥 − ) 13 2 13 (3)-(2): 13𝑦 = 9 13𝑥 1 18 𝑦=− + + 9 1 26 26 26 𝑦= and 𝑥 = 13 13 26y+13x=19 27 1.5.2 Perpendicular distance Geometrically, the perpendicular distance between two objects is the distance from one to the other, measured along a line that is perpendicular to one or both. The perpendicular distance from point 𝑃(𝑚, 𝑛) to line : 𝐴𝑥 + 𝐵𝑦 + 𝐶 = 0 |𝐴𝑚+𝐵𝑛+𝐶| is given by 𝑑 =. √𝐴2 +𝐵2 Example 1.16 Find the perpendicular distance from the line 3𝑦 = 4𝑥 − 1 to the points. a. (1,3) b. (1, −2) Solution: |𝐴𝑚+𝐵𝑛+𝐶| |−4(1)+3(3)+1| 6 |𝐴𝑚+𝐵𝑛+𝐶| |−4(1)+3(−2)+1| 9 𝑑= = = 𝑑= = = √𝐴2 +𝐵2 √(−4)2 +32 5 √𝐴2 +𝐵2 √(−4)2 +32 5 Example 1.17 In each of the following find the perpendicular distance from the given point to the given line. a. 4𝑦 + 3𝑥 + 6 = 0, (2, −1) b. 12𝑦 = 5𝑥 − 1, (−3, −2) Solution: |𝐴𝑚+𝐵𝑛+𝐶| |3(2)+4(−1)+6| 8 |𝐴𝑚+𝐵𝑛+𝐶| |−5(−3)+12(−2)+1| 8 𝑑= = = 𝑑= = = √𝐴2 +𝐵2 √32 +4 2 5 √𝐴2 +𝐵2 √(−5)2 +122 13 28 EXERCISE 1.15 Find which two of the points 𝐴(1,1), 𝐵(0, −1)and 𝐶(−1,2)lie on the same side of the line 2 x + 3 y = 1 and find how far the third point is from the line. Solution: EXERCISE 1.16 Determine the possible values of k given that the point (4, k ) is the same distance from 9 x + 8 y + 1 = 0 as (2,5) is from y = 12 x + 2 Solution: 29 EXERCISE 1.17 A points A, B and C have coordinates (7,9), (−1,2) and (2,6) respectively, determine. a. The equation of the straight line through BC b. The length of BC c. The perpendicular distance from A to BC d. The area of triangle ABC. Solution: 30 1.5.3 Angle between two straight lines. Notice that when two lines intersect, one of the two pairs is acute and the other pair is obtuse. The angle between two lines is defined as the smallest of these angles or the acute angle denoted by θ defined as 𝜃 = 𝑚 −𝑚2 tan−1 1+𝑚 1 where 𝑚1 is a slope for 1 𝑚2 line AB and 𝑚2 is a slope for line CB. Example 1.18 Determine the acute angle between the pair of line whose equation are 3𝑦 = 𝑥 − 7 and 2𝑦 = 3 − 4𝑥. Solution: 1 S1: From 3𝑦 = 𝑥 − 7, 𝑚1 = 3 From 2𝑦 = 3 − 4𝑥, 𝑚2 = −2 S2: Calculate the acute angle, 𝑚1 −𝑚2 𝜃 = tan−1 1+𝑚1 𝑚2 1 −2 3 = tan−1 1 1+( )(−2) 3 = 81.8690 = 890 52′ 31 EXERCISE 1.18 Find the acute angle between the following pairs of lines: a. 𝑦 = 2𝑥 − 3 and 𝑦 = 𝑥 + 4 b. 4𝑦 = 3𝑥 − 4 and 𝑦 + 2𝑥 = 3 Solution: EXERCISE 1.19 Calculate the angles of the triangle ABC where A, B and C are the points (−2,2), (2,4) and (7, −1) respectively. (Hint: sketch the triangle first to determine whether any of the angles are obtuse). Solution: 32 EXERCISE 1.20 Determine the equation of the straight lines that make an angle of 450 with the line 𝑦 = 3𝑥 − 2 and that pass through the point(6,4). Solution: 1.5.4 Equation of bisectors Let us assume the two given straight lines be PQ and RS whose equations are 𝑎1 𝑥 + 𝑏1 𝑦 + 𝑐1 = 0 and 𝑎2 𝑥 + 𝑏2 𝑦 + 𝑐2 = 0 respectively, where 𝑐1 and 𝑐2 are of the symbols. Therefore, the equations of the bisectors of the angles between the lines PQ and RS are denoted as; 𝑎1 𝑥 + 𝑏1 𝑦 + 𝑐1 𝑎2 𝑥 + 𝑏2 𝑦 + 𝑐2 =± √𝑎1 2 + 𝑏1 2 √𝑎2 2 + 𝑏2 2 33 Example 1.19 Find the equations of the bisectors of the angles between the straight lines 4𝑥 − 3𝑦 + 4 = 0 and 6𝑥 + 8𝑦 − 9 = 0. Solution: The equations of the bisectors of the angles between 4𝑥 − 3𝑦 + 4 = 0 and 6𝑥 + 8𝑦 − 9 = 0 are: 4x − 3y + 4 6x + 8y − 9 =± √42 + (−3)2 √62 + 82 4x − 3y + 4 6x + 8y − 9 =± 5 10 40𝑥 − 30𝑦 + 40 = ±(30𝑥 + 40𝑦 − 45) Take positive sign: 40𝑥 – 30𝑦 + 40 = +(30𝑥 + 40𝑦 – 45) ⇒ 2𝑥 − 14𝑦 + 17 = 0 Taking negative sign: 40𝑥 – 30𝑦 + 40 = −(30𝑥 + 40𝑦 – 45) ⇒ 40𝑥 − 30𝑦 + 40 = −30𝑥 − 40𝑦 + 45 ⇒ 70𝑥 + 10𝑦 − 5 = 0 Therefore, the equations of the bisectors of the angles between the straight lines 4𝑥 − 3𝑦 + 4 = 0 and 6𝑥 + 8𝑦 − 9 = 0 are 2𝑥 − 14𝑦 + 17 = 0 and 70𝑥 + 10𝑦 − 5 = 0. 34 Example 1.20 Find the equation of the obtuse angle bisector of lines 4𝑥 − 3𝑦 + 10 = 0 and 8𝑦 − 6𝑥 − 5 = 0. Solution: Make the constant terms positive for both equations: 4𝑥 − 3𝑦 + 10 = 0 and 6𝑥 − 8𝑦 + 5 = 0. Compute 𝑎1 𝑎2 + 𝑏1 𝑏2 = 4 × 6 + (-3) × (-8) = 24 + 24 = 48, which is positive that gives the obtuse angle bisector. The obtuse angle bisector is 4x − 3y + 10 6x − 8y + 5 ⇒ =+ √4 2 +(−3)2 √62 +(−8)2 4x − 3y + 10 6x − 8y+5 ⇒ 5 =+ 10 ⇒ 40𝑥 − 30𝑦 + 100 = 30𝑥 − 40𝑦 − 50 ⇒ 10𝑥 + 10𝑦 + 150 = 0 𝑥 + 𝑦 + 15 = 0, which is the required obtuse angle bisector. EXERCISE 1.21 Find the equations of the lines that bisect the angles between the lines. a. 3𝑥 − 4𝑦 + 13 = 0 and 12𝑥 + 5𝑦 − 32 = 0. b. 𝑦 = 𝑥 + 4 and 7𝑥 + 𝑦 + 3 = 0 Solution: 35 EXERCISE 1.22 Find the locus of the points that are equidistance from the two lines 6𝑦 = 7𝑥 + 1 and 9𝑥 + 2𝑦 + 3 = 0 Solution: 36 Example 1.21 Figure shows a quadrilateral ABCD with coordinate of points A and B is (2, 8) and (8, 6). Given point C lies on perpendicular bisector of line AB, point D is on y-axis, line BC is represented by equation 4x - 3y - 14 = 0, and the angle of DAB is 90o. Determine: a. Equation of line AD y A b. Coordinate of point D B c. perpendicular bisector equation of AB d. Coordinate of point C D x e. Area of quadrilateral ABCD C Solution: a. Equation of line AB 𝑦−8 6−8 = 𝑥−2 8−2 6𝑦 − 48 = −2𝑥 + 2 6𝑦 = −2𝑥 + 50 1 25 𝑦 = −3𝑥 + 3 Equation of line AD: m = 3, A (2, 8) y - 8 = 3(x - 2) y = 3x + 2 37 b. D lies on y-axis, so x = 0 Substitute x = 0 in equation of line AD, y = 2 Coordinate of point D is (0, 2) 2+8 8+6 c. Midpoint = ( , ) = (5, 7), m = 3 2 2 Equation: y - 7 = 3 (x - 5) y = 3x - 8 d. C is intersection point of 4x - 3y = 14 ----(1) 3x - y = 8 -------(2) (2) x 3: 9x - 3y = 24 ----(3) (3) - (1): 5x = 10 → x = 2 and y = -2 Therefore, coordinate C is (2, -2) 12 8 0 2 2 e. Area of quadrilateral ABCD = 2 8 6 2 −2 8 1 = 2 |(12 + 16 + 0 + 16) − (−4 + 4 + 0 + 64)| = 10 unit2 38 Student’s Name: Matric no.: Course name/code: Date of submission: TUTORIAL 1 1. In Figure 1, three points have coordinates 𝐴 (1, 7) , 𝐵 (7, 5) and 1 𝐶(2, −5). The line through C with gradient 2 intersect the line AB at point D. Figure 1 i. Determine the equation of the line AB. ii. Determine the equation of the line passes through C with 1 gradient 2. iii. Show that the point D is (16, 2). 39  iv. If B divides the line AD in the ratio m : n, calculate the values of m and n. v. Calculate the area of triangle ACD. 2. Figure 2 shows a rectangle ABCP. Given that P (x, y) and A (5, 2). The equation of the straight-line AB, 2𝑥 − 3𝑦 − 4 = 0, intersects at point L with AL = BL. The perpendicular bisector of AB intersects the y-axis at point M. Determine: i. The coordinate of point B ii. The equation of the straight-line LM. iii. The coordinate of point M iv. The coordinate of point C, if A, M and C are collinear with the ratio 2:1 v. The perpendicular length from B to the straight-line AC. Figure 2 40