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PQR is a triangle with P(3,4) and Q(9,-2) - Scottish Highers Maths - Question 5 - 2018
Question 5
PQR is a triangle with P(3,4) and Q(9,-2). (a) Find the equation of L1, the perpendicular bisector of PQ. The equation of L2, the perpendicular bisector of PR is 3... show full transcript
Worked Solution & Example Answer:PQR is a triangle with P(3,4) and Q(9,-2) - Scottish Highers Maths - Question 5 - 2018
Step 1
Find the equation of L1, the perpendicular bisector of PQ.
Answer
To find the equation of L1, we first determine the midpoint of PQ, which is given by:
Midpoint = ( M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{3 + 9}{2}, \frac{4 + (-2)}{2} \right) = (6, 1) )
Next, we calculate the slope of line segment PQ, ( m_{PQ} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-2 - 4}{9 - 3} = \frac{-6}{6} = -1 )
The slope of the perpendicular bisector L1 is the negative reciprocal of m_{PQ}, which is ( m_{L1} = 1 ).
Using point-slope form, the equation of L1 is: [ y - 1 = 1(x - 6) ] Simplifying gives: [ y = x - 5 ]
Step 2
Calculate the coordinates of C, the point of intersection of L1 and L2.
Answer
The equation of L2 is given as ( 3y + x = 25 ). To find the intersection C of L1 and L2, we solve the system of equations produced by: [ y = x - 5 ] [ 3y + x = 25 ]
Substituting the expression for y into the equation for L2: [ 3(x - 5) + x = 25 ] [ 3x - 15 + x = 25 ] [ 4x - 15 = 25 ] [ 4x = 40 ] [ x = 10 ]
Now substitute x back to find y: [ y = 10 - 5 = 5 ]
Thus, the coordinates of C are ( C(10, 5) ).
Step 3
Determine the equation of this circle.
Answer
The center of the circle is C(10, 5), and we need to calculate the radius. The radius can be found by measuring the distance from C to any vertex of the triangle (using P or Q).
Using point P(3, 4), the radius r is calculated as follows: [ r = \sqrt{(x_C - x_P)^2 + (y_C - y_P)^2} = \sqrt{(10 - 3)^2 + (5 - 4)^2} = \sqrt{7^2 + 1^2} = \sqrt{49 + 1} = \sqrt{50} ]
Thus, the equation of the circle is: [ (x - 10)^2 + (y - 5)^2 = 50 ]