(a)find the equation of the line which pass through the point (2,-1) and (1,-3).
(b)p(2,1) and Q(1,2)are point in the plane. find the bearing of P and Q
(a) the slope is 2, so use the point-slope form
y+1 = 2(x-2)
(b) Plot the points, and it is easy to see that P is SE of Q.
The bearing of P from Q would be 135°
(2, -1), (1, -3).
a. m = (-3+1)/(1-2) = 2.
y = mx+b.
-1 = 2*2+b
b = -5.
y = 2x-5.
b. P(2, 1), Q(1, 2).
m = Tan A = (1-2)/(2-1) = -1.
A = -45o = 315o CW from +y-axis(Bearing of P and Q).
A = 135o CCW from +x-axis.
(a) To find the equation of a line that passes through two given points, you can use the slope-intercept form of a line, which is y = mx + b, where m is the slope of the line and b is the y-intercept.
To find the slope (m), you can use the formula: m = (y2 - y1) / (x2 - x1), where (x1, y1) and (x2, y2) are the coordinates of the two points.
Using the given points (2, -1) and (1, -3):
m = (-3 - (-1)) / (1 - 2) = -2 / -1 = 2
Now that we have the slope, we can use the point-slope form of a line, which is y - y1 = m(x - x1), to find the equation of the line. Let's use the point (2, -1):
y - (-1) = 2(x - 2)
y + 1 = 2(x - 2)
y + 1 = 2x - 4
y = 2x - 5
Therefore, the equation of the line passing through the points (2, -1) and (1, -3) is y = 2x - 5.
(b) To find the bearing of two points, we need to consider the angle they make with respect to the north (or any reference direction).
To find the bearing between two points (P and Q) in a plane, follow these steps:
1. Determine the change in the y-coordinate (Δy = y2 - y1) and the change in the x-coordinate (Δx = x2 - x1).
2. Calculate the angle using the inverse tangent (arctan) function: bearing = arctan(Δy / Δx).
3. Convert the result from radians to degrees.
Using the points P(2, 1) and Q(1, 2):
Δy = 2 - 1 = 1
Δx = 1 - 2 = -1
bearing = arctan(1 / -1)
bearing ≈ -45 degrees
Therefore, the bearing of point P and Q is approximately -45 degrees.