1. It is known that the line L1 is perpendicular to the line L2: 3x-4y-7 = 0, and the circumference of the triangle formed by L1 and the two coordinate axes is 10. The equation for finding L1 is 2. It is known that the fixed point a (a, 0) is on the straight line x = B (0)

The answer to the first question is the direction vector of L2 (4,3) and the normal vector of L1 n = (4,3). I don't understand this sentence. Is there a problem? How can I see the value of P in the second question? There is still a solution of A. can I confirm it

If the ordinate of the intersection of the line y = KX + B and the line y = 12x + 3 is 5, and the abscissa of the intersection of the line y = 3x-9 is 5, then the area of the triangle formed by the line y = KX + B and the two coordinate axes is ()
A. 32B. 52C. 1D. 12

Substituting y = 5 into y = 12x + 3 to get 12x + 3 = 5, the solution is x = 4, that is, the intersection coordinates of y = KX + B and y = 12x + 3 are (4, 5); substituting x = 5 into y = 3x-9 to get y = 6, that is, the intersection coordinates of y = KX + B and y = 3x-9 are (5, 6); substituting (4, 5) and (5, 6) into y = KX + B to get 4K + B = 55k + B = 6, the solution is k = 1b = 1, so y = x + 1, when x = 0, y = 1; when y = 0, x + 1 = 0, the solution is x = - 1, So the intersection coordinates of the line y = x + 1 and X axis and Y axis are (- 1,0), (0,1), so the area of the triangle formed by the line y = x + 1 and two coordinate axes is 12 × 1 × 1 = 12

If the ordinate of the intersection of the line y = KX + B and the line y = 12x + 3 is 5, and the abscissa of the intersection of the line y = 3x-9 is 5, then the area of the triangle formed by the line y = KX + B and the two coordinate axes is ()
A. 32B. 52C. 1D. 12

Make a straight line L through the point (- 5. - 4) so that it intersects the two coordinate axes and the area of the triangle enclosed by the two axes is 5,

1) One way
Line L, slope k, point (- 5. - 4), K

Given two points P1 and P2, how to find the slope of the straight line P1 and P2

k=(y2-y1)/(x2-x1)

Judge whether the slope of straight line p1p2 exists. If it exists, request its values (1) p1 (1, - 1), P2 (- 3,2) (2) p1 (1, - 2), P2 (5, - 2)

k=(y2-y1)/(x2-x1)
The first k = - 3 / 4
The second k = 0

As for the inclination angle and slope of a straight line, know the slope and the abscissa or ordinate of points P1 and P2, and how to find / p1p2/

Take p1p2 as the hypotenuse, the difference of abscissa and ordinate between P1 and P2 as the right angle, draw a right triangle
The slope is the tangent of the angle opposite the difference in ordinate,
According to this value, we can find the corresponding sine sin a or cosine cos a
Sin a = ordinate difference / ip1p2i, cos a = abscissa difference / ip1p2i
That's enough
According to cosine and tangent or sine and tangent conversion relationship, we can get the conclusion
IP1P2I=│x1-x2│√(k^2+1)=│y1-y2│√[(1/k^2)+1]
Where k is the slope and the magnitude is Tan a
Add the transformation relation (k ^ 2 + 1) = Tan & # 178; a + 1 = Sin & # 178; a / cos & # 178; a + 1 = (Sin & # 178; a + cos & # 178; a) / cos & # 178; a = 1 / cos & # 178; a
IP1P2I=│x1-x2│√(k^2+1)=│x1-x2│√(1/cos²a)=(1/cosa)│x1-x2│
cosa=│x1-x2│/ IP1P2I

Given that the line L passes through two points P1 (x1, Y1) P2 (X2, Y2), is the slope k = y2-y1 / x2-x1 correct

The slope of a straight line is the tangent of the angle between the line and the x-axis, which is equal to the ratio of the difference between the ordinate and abscissa of any two points passing through the line

Given two points P1 (x1, x2) and P2 (X2, Y2), when x1 ≠ X2, the slope formula of line p1p2 is_ When x 1 = x 2, straight
Given two points P1 (x1, x2) and P2 (X2, Y2), when x1 ≠ X2, the slope formula of line p1p2 is When X1 = X2, the inclination angle of line p1p2 is_ When Y1 = Y2, the inclination angle of line p1p2 is_____ .

Two points P1 (x1, x2) and P2 (X2, Y2) are known,
When x1 ≠ X2, the slope formula of line p1p2 is [k = (y2-y1) / (x2-x1)]______ ,
When X1 = X2, the inclination angle of line p1p2 is 90 & # 186_____ ,
When Y1 = Y2, the inclination angle of line p1p2 is_ 0º______ .

1. It is known that the line L1 is perpendicular to the line L2: 3x-4y-7 = 0, and the circumference of the triangle formed by L1 and the two coordinate axes is 10. The equation for finding L1 is 2. It is known that the fixed point a (a, 0) is on the straight line x = B (0)