QUESTION ONE (30 MARKS) a. Find the equation of a line through the point (5,-3) which is inclined at (pi )/(3) to the positive direction of the x-axis. (3marks) b. Find the rectangular co-ordinates of a point B(4,(pi )/(2)) (3marks) C. Change x^2+y^2-4y=0 to polar form. (3marks) d. Find a vector equation and the cartesian equation of the perpendicular bisectors of PQ, where P and Q are the points with position vectors: i) -3i-j and 7i+j (4marks) ii)ai+bj and 2ai+3bj (3marks) e. If z_(1)=2(cos320^circ +jsin320^circ ) and z_(2)=2(cos120^circ +jsin120^circ ) Determine modulus and argument of z_(2)z_(1) (6marks) f.Given A = A=[} 1&1 3&4&5 3&6&5 ] Find i) vert B-Avert (4marks) ii) A X B (4marks) QUESTION TWO (20 MARKS) a. Find an equation of the line parallel to the line with equation 6x-2y=8 and which passes through the point (2,-3) (3marks) b. i) Find a vector equation of the line passing through the points A(2,-2,-1) and B(4,-3,1) (4marks) ii) Hence find the Cartesian equation of the line AB. (3marks)

QUESTION ONE (30 MARKS)a. To find the equation of a line through the point which is inclined at to the positive direction of the x-axis, we can use the point-slope form of a line equation: where is the given point and is the slope of the line. The slope of the line can be found using the tangent function: Substituting the values, we get: Simplifying, we have: b. To find the rectangular coordinates of a point , we can use the polar to Cartesian coordinate conversion formulas: where is the radius and is the angle in radians. Substituting the given values, we get: Therefore, the rectangular coordinates of point B are .c. To change the equation to polar form, we can use the relationships between Cartesian and polar coordinates: Substituting these into the given equation, we get: Simplifying, we have: Dividing both sides by , we get: Therefore, the polar form of the given equation is .d. i) To find the vector equation and the Cartesian equation of the perpendicular bisector of PQ, where P and Q have position vectors and , we can use the formula for the midpoint of a line segment:Midpoint = Substituting the given vectors, we get:Midpoint = The vector equation of the perpendicular bisector is: where is a unit vector perpendicular to PQ. The Cartesian equation can be found by substituting the vector equation into the equation of a plane.ii) For the points and , the same process can be followed to find the vector equation and the Cartesian equation of the perpendicular bisector.e. To determine the modulus and argument of , we can first find the product of and : Using the angle addition formula for cosine and sine, we can simplify the expression: Since is equivalent to , we can rewrite the expression as: The modulus of is given by: The argument of is given by: Therefore, the modulus of is $4\sqrt{