Given ΔABC with A(–3, 2), B(–1, –4), and C(4, 1), write the equation for the line containing altitude in standard form.%0D%0A(1 point)%0D%0AResponses%0D%0A%0D%0Ax – y = 1%0D%0Ax – y = 1%0D%0A%0D%0Ax + y = –1%0D%0Ax + y = –1%0D%0A%0D%0Ax + y = 1%0D%0Ax + y = 1%0D%0A%0D%0Ax – y = –1
To find the equation for the line containing the altitude, we need to first find the slope of the line perpendicular to the base of the triangle and passes through point A.
The slope of the base of the triangle can be found using the coordinates of points B and C:
Slope of base = (y2 - y1) / (x2 - x1)
= (1 - (-4)) / (4 - (-1))
= 5 / 5
= 1
The slope of the line perpendicular to the base is the negative reciprocal of 1, which is -1.
Now, we have the slope of the perpendicular line (-1) and the point A (-3, 2). We can use the point-slope form of a linear equation to write the equation of the line containing the altitude:
y - y1 = m(x - x1)
Plugging in the values, we get:
y - 2 = -1(x - (-3))
y - 2 = -(x + 3)
y - 2 = -x - 3
x + y = 5
Hence, the equation for the line containing the altitude is x + y = 5, which is equivalent to x - y = -5 in standard form.