To prove that the coordinates of the midpoint M on line AC are (0, 1), we first need to find the coordinates of point M.
The coordinates of the midpoint M of a line segment with endpoints A(x1, y1) and C(x2, y2) are given by the formula:
M((x1 + x2)/2, (y1 + y2)/2)
In this case, the coordinates of A are (-2, 3) and the coordinates of C are (2, -1). Plugging these values into the formula, we get:
M(((-2) + 2)/2, (3 + (-1))/2)
M(0, 2/2)
M(0, 1)
Therefore, the coordinates of the midpoint M on line AC are indeed (0, 1).
Next, we need to determine the equation of the line MN passing through M, which is parallel to CB. Since CB has a slope of (2 - (-1))/(5 - 2) = 3/3 = 1, the line MN will also have a slope of 1 in order to be parallel to CB.
The equation of a line with slope m passing through a point (x1, y1) is given by the point-slope formula:
y - y1 = m(x - x1)
Plugging in the values m = 1, x1 = 0, and y1 = 1, we get:
y - 1 = 1(x - 0)
y - 1 = x
y = x + 1
Therefore, the equation of the line MN passing through M and parallel to CB is y = x + 1.
To determine whether the midpoint of AB lies on the line MN, we need to calculate the coordinates of the midpoint of AB and see if they satisfy the equation of the line MN.
The coordinates of the midpoint of AB can be found using the midpoint formula:
M((x1 + x2)/2, (y1 + y2)/2)
In this case, the coordinates of A are (-2, 3) and the coordinates of B are (5, 2). Plugging these values into the formula, we get:
M((-2 + 5)/2, (3 + 2)/2)
M(3/2, 5/2)
The coordinates of the midpoint of AB are (3/2, 5/2). To see if this point lies on the line MN, we substitute x = 3/2 into the equation y = x + 1:
y = 3/2 + 1
y = 5/2
Therefore, the midpoint of AB does lie on the line MN.
To calculate the size of angle O to the nearest degree, we use the slope of line AC, which is given by (3 - (-1))/(-2 - 2) = 4/-4 = -1. The angle O is equal to the arctan of the absolute value of the slope, so:
O = arctan(1) = 45 degrees
To prove that triangle ABC is right-angled, we notice that the slope of line AC is -1, and the slope of line CB is 1. Since these two slopes are negative reciprocals of each other, lines AC and CB are perpendicular and angle C is a right angle. Therefore, triangle ABC is right-angled.
Lastly, to calculate the area of triangle ABC, we can use the formula for the area of a triangle given its vertices:
Area = 1/2 * |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|
Area = 1/2 * |-2(-1 - 2) + 5(3 - (-1)) + 2(2 - 3)|
Area = 1/2 * |6 + 5 + (-2)|
Area = 1/2 * 9
Area = 4.5
Therefore, the area of triangle ABC is 4.5.
Prove the coordinates of the midpoint M on line AC are (0; 1).
1.5 Hence, determine the equation of the line MN passing through M,
Determine whether the midpoint of AB lies on the line MN.
which is parallel to CB.
Calculate the size of 0 to the nearest degree.
Prove thatABC is right-angled.
Calculate the area ofABC.
Figure 1: Diagram for question 1.
A(- 2;3), B(5; 2) and C(2; - 1) are vertices ofABC in a Cartesian plane as shown
below. 0 is the angle of inclination of AC.
1 answer