17. In a triangle ABC has vertices A( 3 , 5 ) , B( 2 , 3 ) , and C( 5 , 2 ). Find the equation of the altitude from A to BC.
18. Find the equation of the median from vertex A in Triangle ABC, if the coordinates of the vertices are
A( -3 , -1 ) , B( 3 , 5 ) , and C( 7 , -3 ).
19. Find the equation of the perpendicular bisector of the line segment joining P( -1 , 4 ) to Q( 3 , -2 ).
18. Find the equation of the median from vertex A in Triangle ABC, if the coordinates of the vertices are
A( -3 , -1 ) , B( 3 , 5 ) , and C( 7 , -3 ).
19. Find the equation of the perpendicular bisector of the line segment joining P( -1 , 4 ) to Q( 3 , -2 ).
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17. We need to find the equation of the line through A(3, 5) that is perpendicular to the line through B( 2 , 3 ) and C( 5 , 2 ).
Slope of line BC = (2-3)/(5-2) = -1/3.
Since the slopes of perpendicular lines are negative reciprocals of each other, the altitude has slope 3 in addition to passing through (3, 5).
From using point-slope, we conclude that the equation of the altitude is
y - 5 = 3(x - 3)
y - 5 = 3x - 9
y = 3x - 4.
18. We need to find the equation of the line through A(-3, -1) to the midpoint of the segment with endpoints B( 3 , 5 ) and C( 7 , -3 ).
This midpoint is ((3+7)/2, (5-3)/2) = (5, 1).
So the median passes through the points (5, 1) and (-3, -1).
The slope of the median is (-1-1)/(-3-5) = 1/4.
From using point-slope, for example on the point (5, 1), we conclude that the equation of the median is
y - 1 = (1/4)(x - 5)
y - 1 = (1/4)x - 5/4
y = (1/4)x - 1/4.
19. We need to find the equation of the line perpendicular to and passing through the midpoint of the segment with endpoints P( -1 , 4 ) and Q( 3 , -2 ).
Slope of segment PQ = (-2-4)/(3+1) = -3/2.
Midpoint of segment PQ = ((-1+3)/2, (4-2)/2) = (1, 1).
Since the slopes of perpendicular lines are negative reciprocals of each other, the perpendicular bisector has slope 2/3.
The perpendicular bisector also passes through the midpoint (1, 1).
From using point-slope, we conclude that the equation of the perpendicular bisector is
y - 1 = (2/3)(x - 1)
y - 1 = (2/3)x - 2/3
y = (2/3)x + 1/3.
Lord bless you today!
Slope of line BC = (2-3)/(5-2) = -1/3.
Since the slopes of perpendicular lines are negative reciprocals of each other, the altitude has slope 3 in addition to passing through (3, 5).
From using point-slope, we conclude that the equation of the altitude is
y - 5 = 3(x - 3)
y - 5 = 3x - 9
y = 3x - 4.
18. We need to find the equation of the line through A(-3, -1) to the midpoint of the segment with endpoints B( 3 , 5 ) and C( 7 , -3 ).
This midpoint is ((3+7)/2, (5-3)/2) = (5, 1).
So the median passes through the points (5, 1) and (-3, -1).
The slope of the median is (-1-1)/(-3-5) = 1/4.
From using point-slope, for example on the point (5, 1), we conclude that the equation of the median is
y - 1 = (1/4)(x - 5)
y - 1 = (1/4)x - 5/4
y = (1/4)x - 1/4.
19. We need to find the equation of the line perpendicular to and passing through the midpoint of the segment with endpoints P( -1 , 4 ) and Q( 3 , -2 ).
Slope of segment PQ = (-2-4)/(3+1) = -3/2.
Midpoint of segment PQ = ((-1+3)/2, (4-2)/2) = (1, 1).
Since the slopes of perpendicular lines are negative reciprocals of each other, the perpendicular bisector has slope 2/3.
The perpendicular bisector also passes through the midpoint (1, 1).
From using point-slope, we conclude that the equation of the perpendicular bisector is
y - 1 = (2/3)(x - 1)
y - 1 = (2/3)x - 2/3
y = (2/3)x + 1/3.
Lord bless you today!