Two straight lines L1: MX + 8y + n = 0 and L2: 2x + MY-1 = 0 are known. Try to determine the value of M, n so that (1) L1 ‖ L2; (2) L1 ⊥ L2, and the intercept of L1 on the Y axis is - 1

Two straight lines L1: MX + 8y + n = 0 and L2: 2x + MY-1 = 0 are known. Try to determine the value of M, n so that (1) L1 ‖ L2; (2) L1 ⊥ L2, and the intercept of L1 on the Y axis is - 1

(1) When m = 0, it is obvious that L1 and L2 are not parallel. & nbsp; when m ≠ 0, from M2 = 8m ≠ n-1 & nbsp; we get m · M-8 × 2 = 0, M = ± 4,8 × (- 1) - n · m ≠ 0, n ≠± 2, so when m = 4, n ≠ - 2, or M = - 4, n ≠ 2, L1 ‖ L2. (2) if and only if M · 2 + 8 · M = 0, that is, when m = 0, L1 ‖ L2

If a line L passes through a point (1,2) and its intercept on the y-axis is 1, then the equation of the line L is? How do I start with this question?

The ladder process of this question
We must know that the linear equation is y = KX + B
The intercept on the Y axis is 1, which means that when x = 0, y = 1, so B = 1
When x = 1, y = 2
By introducing y = KX + 1, we get k = 1
So the equation is y = x + 1

If the line L passes through the point (- 4, - 1) and the transverse intercept is twice of the longitudinal intercept, then the general equation of the line L is______ ．

(1) When the intercept of the straight line is 0, the equation of the straight line passing through the origin (0, 0) is y = 14x, which can be changed into the general formula to get x-4y = 0; (2) when the intercept of the straight line is not 0, let the equation be x2a + ya = 1, substituting (- 4, - 1) into the point to get − 42a + − 1A = 1, and the solution is a = - 3

The intercept of point P (2, - 1) on x-axis and y-axis is a and B respectively, which satisfies the linear equation of a = 3B

If a = 0, the linear equation is y = - 12x; if a ≠ 0, let the linear equation be XA + Yb = 1, then a = - 1, B = - 13, and the obtained linear equation is x + 2Y = 0 or x + 3Y + 1 = 0