© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the coordinates of the given points
The triangle has vertices
A( -2, 3 ), B( 1, 9 ), and C( 3, 8 ).
Step 2: Find the midpoint of BC
The midpoint M of BC is given by
M \bigl(\frac{x_B + x_C}{2}, \;\frac{y_B + y_C}{2}\bigr).
Substituting B( 1,\,9 ) and C( 3,\,8 ):
M \Bigl(\frac{1+3}{2},\,\frac{9+8}{2}\Bigr)
= \bigl(2,\,\frac{17}{2}\bigr).
Step 3: Find the circumcentre of triangle ABC
The circumcentre is the intersection of the perpendicular bisectors of any two sides. We will use sides AB and AC.
(a) Perpendicular bisector of AB
Midpoint of AB:
\Bigl(\frac{-2 + 1}{2},\,\frac{3 + 9}{2}\Bigr)
= \Bigl(-\frac{1}{2},\,6\Bigr).
Slope of AB:
m_{AB} = \frac{9 - 3}{\,1 - (-2)\,} = \frac{6}{3} = 2.
Hence, the slope of the perpendicular bisector is
-\frac{1}{2} .
Equation of the perpendicular bisector of AB passing through
(-\tfrac{1}{2},\,6) :
y - 6 \;=\; -\tfrac{1}{2}\,\bigl(x + \tfrac{1}{2}\bigr).
Simplifying,
y = -\tfrac{1}{2}x + \tfrac{23}{4}.
(b) Perpendicular bisector of AC
Midpoint of AC:
\Bigl(\frac{-2 + 3}{2},\,\frac{3 + 8}{2}\Bigr)
= \Bigl(\frac{1}{2},\,\frac{11}{2}\Bigr).
Slope of AC:
m_{AC} = \frac{8 - 3}{\,3 - (-2)\,} = \frac{5}{5} = 1.
Hence, the slope of the perpendicular bisector is
-1 .
Equation of the perpendicular bisector of AC passing through
(\tfrac{1}{2},\,\tfrac{11}{2}) :
y - \tfrac{11}{2} = -1\bigl(x - \tfrac{1}{2}\bigr).
Simplifying,
y = -x + 6.
(c) Intersection of the two perpendicular bisectors
The circumcentre is the point of intersection of
y = -\tfrac{1}{2}x + \tfrac{23}{4}
and
y = -x + 6 .
Set them equal:
-\tfrac{1}{2}x + \tfrac{23}{4} \;=\; -x + 6.
Multiply through by 4:
-2x + 23 \;=\; -4x + 24.
Rearrange:
2x = 1
\quad\Longrightarrow\quad
x = \tfrac{1}{2}.
Substituting x = \tfrac{1}{2} into y = -x + 6 :
y = -\tfrac{1}{2} + 6 = \tfrac{11}{2}.
Thus, the circumcentre is
O\bigl(\tfrac{1}{2},\,\tfrac{11}{2}\bigr).
Step 4: Equation of line L passing through the circumcentre and midpoint of BC
The line L goes through
O(\tfrac{1}{2},\,\tfrac{11}{2})
and
M(2,\,\tfrac{17}{2}) .
Compute the slope of OM:
m = \frac{\tfrac{17}{2} \;-\; \tfrac{11}{2}}{\,2 - \tfrac{1}{2}\,}
= \frac{6/2}{3/2}
= \frac{3}{\,\tfrac{3}{2}\,}
= 2.
Equation of the line with slope 2 through O(\tfrac{1}{2},\,\tfrac{11}{2}) :
y - \tfrac{11}{2} = 2\Bigl(x - \tfrac{1}{2}\Bigr).
Simplifying,
y - \tfrac{11}{2} = 2x - 1
\quad\Longrightarrow\quad
y = 2x - 1 + \tfrac{11}{2}
= 2x + \tfrac{9}{2}.
Step 5: Find intersection of line L with the y-axis
For the y-axis, x = 0 . Substituting x=0 into
y = 2x + \frac{9}{2} :
y = \frac{9}{2}.
The intersection with the y-axis is thus
\bigl(0,\,\tfrac{9}{2}\bigr) .
Step 6: Compare with the given form (0,\frac{\alpha}{2})
The problem states that line L intersects the y-axis at
\Bigl(0,\frac{\alpha}{2}\Bigr) . We have found the intersection to be
\bigl(0,\tfrac{9}{2}\bigr) .
Hence,
\frac{\alpha}{2} = \frac{9}{2}
\quad\Longrightarrow\quad
\alpha = 9.
Final Answer
\alpha = 9
