This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Okay, Meendahmwanakashi1, let's go through Question Two completely. Question Two: PQ is the straight line passing through the points P(6,1) and Q(0,-3). a) Find the gradient (m) of the line PQ. Step 1: Use the gradient formula m = (y_2 - y_1)/(x_2 - x_1). Let P be (x_1, y_1) = (6,1) and Q be (x_2, y_2) = (0,-3). m = (-3 - 1)/(0 - 6) Step 2: Calculate the gradient. m = (-4)/(-6) = (2)/(3) The gradient of the line PQ is (2)/(3). b) Write down the equation of the line PQ. Step 1: Use the point-slope form y - y_1 = m(x - x_1) with the gradient m = (2)/(3) and point Q(0,-3). y - (-3) = (2)/(3)(x - 0) Step 2: Simplify the equation. y + 3 = (2)/(3)x Step 3: Isolate y. y = (2)/(3)x - 3 The equation of the line PQ is y = (2)/(3)x - 3. c) Calculate the perpendicular bisector meets PQ intersects at A. This question asks for the point A where the perpendicular bisector of PQ intersects the line segment PQ. This point is the midpoint of PQ. Step 1: Use the midpoint formula A = ((x_1+x_2)/(2), (y_1+y_2)/(2)). Let P be (x_1, y_1) = (6,1) and Q be (x_2, y_2) = (0,-3). A = ((6+0)/(2), (1+(-3))/(2)) Step 2: Calculate the coordinates of A. A = ((6)/(2), (-2)/(2)) = (3, -1) The point A is (3, -1). d) Find the coordinates of A. This is the same as part c). The coordinates of A are (3, -1). e) Write down the equation of the perpendicular line. This refers to the equation of the perpendicular bisector of PQ. Step 1: Find the gradient of the perpendicular bisector. The gradient of PQ is m_PQ = (2)/(3). The gradient of a line perpendicular to PQ, m_, is the negative reciprocal. m_ = -(1)/(m_PQ) = -(1)/(2/3) = -(3)/(2) Step 2: Use the point-slope form y - y_A = m_(x - x_A) with the gradient m_ = -(3)/(2) and the midpoint A(3,-1). y - (-1) = -(3)/(2)(x - 3) Step 3: Simplify the equation. y + 1 = -(3)/(2)x + (9)/(2) Step 4: Isolate y. y = -(3)/(2)x + (9)/(2) - 1 y = -(3)/(2)x + (9)/(2) - (2)/(2) y = -(3)/(2)x + (7)/(2) The equation of the perpendicular line is y = -(3)/(2)x + (7)/(2). f) State any two (2) real life applications of Coordinate Geometry. • Navigation and GPS: Coordinate geometry is fundamental to mapping and global positioning systems, allowing us to locate positions and plan routes. • Engineering and Architecture: It is used in designing structures, buildings, and bridges by defining points, lines, and shapes in a coordinate system. Drop the next question.