Choose the equation of the line passing through the point (-1, -2) and perpendicular to y = one fourthx + 2.
a landing airplane, moving in a straight line , passes over two points (p and q) which are 300 feet apart. when the airplane is above p, as seen from Q, its angle of elevation is 60 degress, and when above Q, as seen from p, its angle of elevation is 40 degress. at what distance from q will the airplane land?
To find the equation of a line that is perpendicular to another line, we need to use the negative reciprocal of the slope of the given line.
Given line: y = (1/4)x + 2
The slope of this line is (1/4). So, the negative reciprocal of (1/4) is -4/1, or -4.
Now, we have the slope (-4) and a point (-1, -2) through which the line passes. We can use the point-slope form of a line to find the equation of the line:
y - y1 = m(x - x1)
where (x1, y1) is the given point and m is the slope.
Plugging in the values, we get:
y - (-2) = -4(x - (-1))
Simplifying:
y + 2 = -4(x + 1)
Expanding the brackets:
y + 2 = -4x - 4
Moving the terms around:
y = -4x - 6
So, the equation of the line passing through the point (-1, -2) and perpendicular to y = (1/4)x + 2 is y = -4x - 6.
you want a slope of -4
y=-4x+b
put in that (x,y) point and solve for b.