Given ΔABC with A(–3, 2), B(–1,
–4), and C(4, 1), write the
equation for the line containing
altitude in standard form.
(1 point)
x – y = 1
x + y = –1
x + y = 1
x – y = –1
To find the equation of the line containing the altitude in standard form, we first need to find the slope of the base AC, by calculating the slope of the line passing through points A and C.
Slope of AC = (y2 - y1) / (x2 - x1)
Slope of AC = (1 - 2) / (4 - (-3))
Slope of AC = -1 / 7
Since the slope of a line perpendicular to another line is the negative reciprocal of the original line's slope, the slope of the altitude is 7/1 = 7.
Now, we can use the point-slope form of a linear equation to find the equation of the line containing the altitude, using point B(–1, –4) through which the altitude passes:
y - y1 = m(x - x1)
y - (-4) = 7(x - (-1))
y + 4 = 7(x + 1)
y + 4 = 7x + 7
y = 7x + 3
Rearranging the equation into standard form (Ax + By = C) gives the final answer:
-7x + y = 3
-7x + y - 3 = 0
Therefore, the equation for the line containing the altitude in standard form is -7x + y - 3 = 0.