Given a (- 1, - 5), B (3, - 2), the inclination angle of line L is twice that of line AB, and the slope of line L is calculated

Given a (- 1, - 5), B (3, - 2), the inclination angle of line L is twice that of line AB, and the slope of line L is calculated

The slope of the line AB is (- 2 + 5) \ (3 + 1) = 3 / 4. Note that the inclination of AB is x and that of L is y. then the slope of 2x = y. the slope of L is tany = tan2x = (2tanx) \ (the square of 1-tanx) = 24 / 7

Given that the slope of the straight line L is √ 3, the inclination angle of the straight line m is twice that of the straight line L, the slope of the straight line M can be calculated
It is known that the three vertices of △ ABC are a (7,8) B (O, 4) C (2, - 4) respectively. Let three lines x-2y + 3 = 0,3x + 4y-21 = 0,2x + 3y-k = 0 be at the same point, then k =?

Solution 1: because the slope of the straight line L is the tangent of the inclination angle of the straight line L, and is √ 3, there is tan60 ° = √ 3
So the inclination angle of the line L is 60 degrees
So the inclination angle of the straight line m is 120 degrees
So the slope of the straight line m is Tan 120 ° = - 3
Solution 2: the abscissa of the midpoint of AB side is (7 + 0) / 2 = 3.5, and the ordinate of the midpoint of AB side is (8 + 4) / 2 = 6
The coordinates of the midpoint of AB side are (3.5,6)
The midline of AB side passes through point (3.5,6) and point C (2, - 4)
According to the two-point formula of the linear equation, the equation of the middle line on the AB side is as follows:
(y-6)/(x-3.5)=(-4-6)/(2-3.5)
The general formula is: 20x-3y-52 = 0
Solve problem 3: three linear equations are set up to form a system of equations, and the intersection coordinates of the three lines are obtained
x-2y+3=0
3x+4y-21=0
2x+3y-k=0
Solve the equations, get k, get: k = 15

If Sina * cosa > 0, which quadrant is a?

∵sinα×cosα>0,
∴sinα0
A is the angle of the first or third quadrant