Why is the equation of a line passing through point a (x0.y0) and parallel to the line ax + BX + C1 equal to 0 generally: a (x minus x0) plus B (y minus Y0) equals zero?

The linear equation parallel to AX + BX + C1 equals 0 must be
Ax + BX + D = 0 and a (x0, Y0)
Ax0+By0+D=0
D = - (ax0 + by0), substituting into the equation AX + BX + D = 0, we get a (x-x0) + B (y-y0) = 0

The linear equation passing through point a (x0, Y0) and parallel to AX + BX + C = 0 is the? Step

If it is parallel to the line, then the slope is equal
Ax+by+D=0
If D is unknown, substitute the point a (x0, Y0) to get the value of D
Is - ax0-by0
So: a (x-x0) + B (y-y0) = 0

Design an algorithm to determine the number of solutions of the quadratic equation AX ^ 2 + BX + C = 0~

In fact, it's very simple. The key is to judge the symbol of the discriminant. It can be divided into three cases: first, the discriminant > = 0 or = 0, and then two kinds of > 0 or x = 0 (using judgment statements), and then output the results respectively. You can draw your own program diagram

Point a (x0, Y0) is on the right branch of hyperbola x ^ 2 / 4-y ^ 2 / 32 = 1. If the distance of point a with right focus is equal to 2x0, then x0 is_____

X ^ 2 / 4-y ^ 2 / 32 = 1A ^ 2 = 4, B ^ 2 = 32, C ^ 2 = 4 + 32 = 36, so the right focus f coordinate is (6,0) AF ^ 2 = (xo-6) ^ 2 + yo ^ 2 = (2xo) ^ 28xo ^ 2-yo ^ 2 = 32xo ^ 2-12xo + 36 + 8xo ^ 2-32 = 4xo ^ 25xo ^ 2-12xo + 4 = 0 (5xo-2) (xo-2) = 0xo = 2 / 5, XO = 2, XO = 2 because XO > = a = 2, so XO = 2

Why is the equation of a line passing through point a (x0.y0) and parallel to the line ax + BX + C1 equal to 0 generally: a (x minus x0) plus B (y minus Y0) equals zero?