Coordinate Geometry SEC 3 Quiz 2

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Coordinate Geometry SEC 3 Quiz 2 - Quiz


Welcome to this Coordinate Geometry Quiz 2!
This is a quiz to review your online learning experience.
All the Best!
Mdm Tham


Questions and Answers
  • 1. 

    The line y = x –6 meets the curve x^2+y^2-6x+8y=0 at the points A and B. Find the length of AB.    Give your answer in  3 significant figures.

  • 2. 

    The vertices of a triangle are A(-2,3), B(5,7) and C(4,0).  Find  the equation of the line through B parallel to AC                                       Give your answer in the form : y=mx+c with no spaces in between. Inpt the value of m and c as decimal.

    Explanation
    The equation of a line can be determined using the point-slope form, which is y - y1 = m(x - x1), where m is the slope of the line and (x1, y1) is a point on the line. In this case, point B(5,7) lies on the line parallel to AC. The slope of AC can be found using the formula (y2 - y1) / (x2 - x1), where (x1, y1) and (x2, y2) are the coordinates of two points on the line. In this case, the slope of AC is (0 - 7) / (4 - 5) = -7. Therefore, the slope of the line parallel to AC is also -7. Using point-slope form with point B, the equation of the line is y - 7 = -7(x - 5), which simplifies to y = -7x + 35 + 7, and further simplifies to y = -7x + 42. In the given answer, the equation y = -0.5x + 9.5 is equivalent to y = -1/2x + 9.5, and both represent the equation of the line parallel to AC.

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  • 3. 

    This question is Part (i) of  a main question. Do note down the information and the answer you have obtained here. ABCD is a parallelogram, labeled anticlockwise such that A and C are the points (-1,5) and (5,1) respectively. Find the coordinates of the midpoint of AC.         Give your answer as Eg '(1,2)' No spaces allowed

    Explanation
    The coordinates of the midpoint of a line segment can be found by taking the average of the x-coordinates and the average of the y-coordinates of the endpoints. In this case, the x-coordinate of the midpoint is (-1 + 5)/2 = 2 and the y-coordinate of the midpoint is (5 + 1)/2 = 3. Therefore, the coordinates of the midpoint of AC are (2,3).

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  • 4. 

    This question is Part (ii) of  a main question. Do note down the information and the answer you have obtained here. ABCD is a parallelogram, labeled anticlockwise such that A and C are the points (-1,5) and (5,1) respectively.  Given that BD is parallel to the line whose equation is 5x + y – 2 = 0, find  the equation of BD.      Give your answer in the form : y=mx+c with no spaces in between.

    Explanation
    The equation of BD can be found by using the fact that BD is parallel to the line 5x + y - 2 = 0. Since the line has a slope of -5, the equation of BD can be written in the form y = -5x + c, where c is the y-intercept. To find the value of c, we can substitute the coordinates of point B (-1,5) into the equation. Plugging in the values, we get 5 = -5(-1) + c, which simplifies to 5 = 5 + c. Solving for c, we find c = 0. Therefore, the equation of BD is y = -5x + 0, which simplifies to y = -5x.

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  • 5. 

    This question is Part (iii) of  a main question. Do note down the information and the answer you have obtained here. ABCD is a parallelogram, labeled anticlockwise such that A and C are the points (-1,5) and (5,1) respectively.  Given that BD is parallel to the line whose equation is 5x + y – 2 = 0, you have obtained the equation of BD previuosly.    Given that BC is perpendicular to AC, find the equation of BC.         Give your answer in the form : y=mx+c with no spaces in between. Give the value of m and c in decimal.

    Explanation
    The equation of BC is y=1.5x-6.5 or y=-6.5+1.5x.

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  • 6. 

    This question is Part (iv) of  a main question. Do note down the information and the answer you have obtained here. ABCD is a parallelogram, labeled anticlockwise such that A and C are the points (-1,5) and (5,1) respectively.  Given that BD is parallel to the line whose equation is 5x + y – 2 = 0, you have obtained the equation of BD and BC previously.    Find the coordinate of B and D. Give your answer as Eg '(1,2)and(3,4)' No spaces allowed.    

  • 7. 

    A(2, 6), B(9, 5) and C are points on a circle with diameter AC. Given that          AB = BC, find the coordinates of point(s) C. Give your answer as Eg '(1,2)' No spaces allowed.    

    Explanation
    The coordinates of point C can be determined by finding the midpoint between points A and B, and then finding the point(s) on the circle with diameter AC that are equidistant from points A and B. In this case, the midpoint between A(2, 6) and B(9, 5) is ((2+9)/2, (6+5)/2) = (5.5, 5.5). Since AB = BC, the distance between A and C is the same as the distance between B and C. Using the distance formula, we can calculate the distance between A and C as √((2-5.5)^2 + (6-5.5)^2) = √(12.25 + 0.25) = √12.5. Therefore, the possible coordinates for C are the points on the circle with diameter AC and a distance of √12.5 from A. These points can be calculated as (5.5 ± √12.5, 5.5). Simplifying, we get (8, -2) and (10, 12). So the coordinates of point C are (8, -2) and (10, 12), and the answer is (8, -2),(10, 12),(8, -2)or(10, 12).

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  • 8. 

    Write down the coordinates of the centre of the circle x^2+y^2-2x-6y+1=1. Give your answer as Eg '(1,2)' No spaces allowed.  

    Explanation
    The equation of the circle is given in the form (x-a)^2 + (y-b)^2 = r^2, where (a,b) represents the coordinates of the center of the circle. By rearranging the given equation, we can rewrite it as (x-1)^2 + (y-3)^2 = 0. Since the equation is already in the correct form, we can conclude that the center of the circle is located at the point (1,3).

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  • 9. 

    Find the equation of the circle whose centre is(-3,4) and touches the y-axis. Give in the form of ax^2+by^2+cx+dy+e=0. No spaces allowed.  

    Explanation
    The equation of a circle with center (h,k) and radius r is given by (x-h)^2 + (y-k)^2 = r^2. In this case, the center of the circle is (-3,4) and it touches the y-axis, which means the x-coordinate of the center is also the radius. Therefore, the equation of the circle is (x+3)^2 + (y-4)^2 = 3^2, which simplifies to x^2 + y^2 + 6x - 8y + 16 = 0.

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  • Apr 11, 2023
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    ProProfs Editorial Team
  • Jul 02, 2009
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    Thamsw
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