Find the gradient, equation of the line PQ, and the coordinates of the intersection point A of the perpendicular bisector with PQ.

Welcome back — been a while! Let's pick up where you left off. QUESTION ONE a) Find gradient (m) of the line PQ. Given points P(-2, -1) and Q(4, 2). The formula for the gradient m is m = (y_2 - y_1)/(x_2 - x_1). Step 1: Substitute the coordinates into the formula. m = (2 - (-1))/(4 - (-2)) Step 2: Simplify the expression. m = (2 + 1)/(4 + 2) = (3)/(6) = (1)/(2) The gradient of line PQ is (1)/(2). b) Write down the equation of the line PQ. Using the point-slope form y - y_1 = m(x - x_1) with m = (1)/(2) and point P(-2, -1). Step 1: Substitute the values. y - (-1) = (1)/(2)(x - (-2)) y + 1 = (1)/(2)(x + 2) Step 2: Simplify to slope-intercept form y = mx + c. y + 1 = (1)/(2)x + 1 y = (1)/(2)x + 1 - 1 y = (1)/(2)x The equation of the line PQ is y = (1)/(2)x. c) Calculate the perpendicular bisector meets PQ intersects at A. d) Find the coordinates of A. Point A is the midpoint of PQ. The formula for the midpoint M is M = ((x_1+x_2)/(2), (y_1+y_2)/(2)). Step 1: Substitute the coordinates of P and Q. A = ((-2+4)/(2), (-1+2)/(2)) Step 2: Calculate the coordinates. A = ((2)/(2), (1)/(2)) = (1, (1)/(2)) The coordinates of A are (1, (1)/(2)). e) Write down the equation of the perpendicular line. This refers to the perpendicular bisector of PQ. Step 1: Find the gradient of the perpendicular bisector. The gradient of PQ is m_PQ = (1)/(2). The gradient of the perpendicular bisector m_ is the negative reciprocal of m_PQ. m_ = -(1)/(m_PQ) = -(1)/(1)2 = -2 Step 2: Use the point-slope form y - y_A = m_(x - x_A) with m_ = -2 and point A(1, (1)/(2)). y - (1)/(2) = -2(x - 1) Step 3: Simplify to slope-intercept form. y - (1)/(2) = -2x + 2 y = -2x + 2 + (1)/(2) y = -2x + (4)/(2) + (1)/(2) y = -2x + (5)/(2) The equation of the perpendicular bisector is y = -2x + (5)/(2). f) State any two (2) real life applications of Coordinate Geometry. • Navigation systems (GPS) use coordinates to pinpoint locations and plan routes. • Engineering and architecture use coordinate geometry for designing structures, mapping out construction sites, and ensuring precise measurements. QUESTION THREE a) Define assessment. Assessment is the systematic process of documenting and using empirical data on knowledge, skills, attitudes, and beliefs to refine programs and improve student learning. It involves gathering information about student performance. b) State two (2) purposes of assessing learners. • To monitor student progress and identify areas where they may need additional support or challenge. • To evaluate the effectiveness of teaching methods and curriculum, providing feedback for instructional improvement. c) State the difference between mathematics and mathematics education. Mathematics is the pure academic discipline focused on the study of quantity, structure, space, and change. Mathematics education is the field concerned with the methods, theories, and practices of teaching and learning mathematics. d) Describe an effective teacher. An effective teacher creates a positive and engaging learning environment, possesses strong subject matter knowledge, uses diverse instructional strategies, and provides constructive feedback to foster student growth. They are also adaptable and responsive to student needs. e) State the difference between a Lesson Plan and Schemes of Work. A Lesson Plan is a detailed, day-to-day guide for a single lesson, outlining specific objectives, activities, materials, and assessment for that particular class period. A Scheme of Work is a broader, long-term plan that outlines the content, learning objectives, and sequence of topics to be covered over an extended period, such as a term or year. QUESTION FOUR a) Simplify: (y^2+4y+3)/(2y^2+7y+5) Step 1: Factorise the numerator. y^2+4y+3 = (y+1)(y+3) Step 2: Factorise the denominator. 2y^2+7y+5 = 2y^2+2y+5y+5 = 2y(y+1)+5(y+1) = (2y+5)(y+1) Step 3: Substitute the factored expressions back into the fraction and simplify. ((y+1)(y+3))/((2y+5)(y+1)) = (y+3)/(2y+5) The simplified expression is (y+3)/(2y+5). b) Express as a single fraction: (2k-2t)/(15) + (k-t)/(3) Step 1: Find a common denominator, which is 15. (2k-2t)/(15) + ((k-t) × 5)/(3 × 5) (2k-2t)/(15) + (5(k-t))/(15) Step 2: Combine the numerators over the common denominator. (2k-2t + 5k-5t)/(15) Step 3: Simplify the numerator. (7k-7t)/(15) The expression as a single fraction is (7k-7t)/(15). c) Solve: (x+4)/(2) - (2x+1)/(3) = 0 Step 1: Find a common denominator for the fractions, which is 6. Multiply the entire equation by 6. 6((x+4)/(2)) - 6((2x+1)/(3)) = 6(0) 3(x+4) - 2(2x+1) = 0 Step 2: Distribute and simplify. 3x + 12 - 4x - 2 = 0 Step 3: Combine like terms. -x + 10 = 0 Step 4: Solve for x. -x = -10 x = 10 The solution is x=10. d) Solve the simultaneous equations: 2x+3y=-4 and x-y=3 Step 1: From the second equation, express x in terms of y. x - y = 3 x = y + 3 Step 2: Substitute this expression for x into the first equation. 2(y+3) + 3y = -4 Step 3: Distribute and solve for y. 2y + 6 + 3y = -4 5y + 6 = -4 5y = -4 - 6 5y = -10 y = -2 Step 4: Substitute the value of y back into x = y+3 to find x. x = -2 + 3 x = 1 The solution is x=1, y=-2. e) Factorise the following expressions; i. x+8x+7 Assuming the expression is x^2+8x+7. We need two numbers that multiply to 7 and add to 8. These numbers are 1 and 7. x^2+8x+7 = (x+1)(x+7) The factorised expression is (x+1)(x+7). ii. 6-5b-4b^2 Step 1: Rearrange the terms in standard quadratic form. -4b^2 - 5b + 6 Step 2: Factor out -1 to make the leading coefficient positive. -(4b^2 + 5b - 6) Step 3: Factor the quadratic expression 4b^2 + 5b - 6. We look for two numbers that multiply to 4 × -6 = -24 and add to 5. These numbers are 8 and -3. 4b^2 + 8b - 3b - 6 4b(b+2) - 3(b+2) (4b-3)(b+2) Step 4: Include the factored -1. -(4b-3)(b+2) This can also be written as (3-4b)(b+2). The factorised expression is (3-4b)(b+2). Send me the next one 📸