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5 The line 2y - x = 12 intersects the circle x+y?-10x –12y + 36 = 0 at the points A and B.

a Find the coordinates of the points A and B.
b Find the equation of the perpendicular bisector of AB.
c The perpendicular bisector of AB intersects the circle at the points P and 0.
Find the exact coordinates of P and Q.
d Find the exact area of quadrilateral APBQ.

User IkiK
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Final answer:

To find the coordinates of points A and B, we solve the system of equations formed by the line and the circle. The equation of the perpendicular bisector of AB is y = -13/4x + 37/4. The coordinates of points P and Q are (1, 10) and (61/4, -6) respectively. The area of quadrilateral APBQ is 64 square units.


Step-by-step explanation:

To find the coordinates of points A and B, we can solve the system of equations formed by the line and the circle. Substituting y = (1/2)x - 6 into the equation of the circle, we get x^2 + (1/2)x - 10x - 6 - 12(1/2)x + 36 = 0. Simplifying this equation gives us x^2 - (19/2)x + 30 = 0. Solving this quadratic equation, we find that x = 2 and x = 15/2. Plugging these x-values into y = (1/2)x - 6, we find that the corresponding y-values are y = -5 and y = 3/2. Therefore, the coordinates of point A are (2, -5) and the coordinates of point B are (15/2, 3/2).

To find the equation of the perpendicular bisector of AB, we first find the midpoint of segment AB. The midpoint is given by the average of the x-coordinates and the average of the y-coordinates, so the midpoint is ((2 + 15/2)/2, (-5 + 3/2)/2), which simplifies to (9/2, -7/4). The slope of the line AB is (3/2 - (-5))/(15/2 - 2) = 4/13. Therefore, the slope of the perpendicular bisector is -13/4. Using the point-slope form of a linear equation, we get y - (-7/4) = -13/4(x - 9/2). Simplifying this equation gives us y = -13/4x + 13/2 + 7/4, which simplifies to y = -13/4x + 37/4.

To find the coordinates of points P and Q, we can solve the system of equations formed by the circle and the equation of the perpendicular bisector. Substituting y = -13/4x + 37/4 into the equation of the circle, we get x^2 + (-13/4x + 37/4) - 10x - 12(-13/4x + 37/4) + 36 = 0. Simplifying this equation gives us x^2 - (47/4)x + 61/4 = 0. Solving this quadratic equation, we find that x = 1 and x = 61/4. Plugging these x-values into y = -13/4x + 37/4, we find that the corresponding y-values are y = 10 and y = -6. Therefore, the coordinates of point P are (1, 10) and the coordinates of point Q are (61/4, -6).

To find the area of quadrilateral APBQ, we can use the Shoelace Formula which states that the area of a quadrilateral with vertices (a, b), (c, d), (e, f), and (g, h) is given by the absolute value of (ac + ce + eg + ga - bd - df - fh - hb)/2. Plugging in the coordinates of A, B, P, and Q into this formula, we get the area to be |(2(20) + (9/2)(-5) + (15/2)(-6) + 1(-7))/2| = |40 - 45/2 - 45 - 7/2| = |-128/2| = 64 square units.


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User Davey Chu
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