Final answer:
The equation of a line perpendicular to BC and passing through the point A(3,8) is y = (7/8)x + (43/8).
Step-by-step explanation:
The question involves finding the equation of a line that is perpendicular to the line BC and passes through the point A(3,8) in the vertices of a ΔABC.
Step-by-Step Solution
First, let's find the slope of the line BC using the coordinates of points B(-1,2) and C(6,-6).
Slope of BC (mBC)= (y2-y1) / (x2-x1) = (-6-2) / (6+1) = -8/7
The slope of line perpendicular to BC (mperpendicular) will be the negative reciprocal of mBC. Hence, mperpendicular = 7/8.
Now, using the point-slope form of the equation of a line which passes through A(3,8) and has a slope of 7/8:
y - y1 = mA(x - x1)
y - 8 = (7/8)(x - 3)
The equation of a line in slope-intercept form (y = mx + b) will then be:
y = (7/8)x - (21/8) + 8
y = (7/8)x + (43/8)
So, the equation of a line perpendicular to BC and passing through point A is y = (7/8)x + (43/8).