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How many math topics are there in the liberal arts class in senior two?
1. In the form of k(x-3)=y- 1, it can be seen that no matter how K changes, the linear beam equation passes through the fixed point (3, 1).

This kind of problem is simplified and sorted into a standard form of point-oblique equation. Sometimes you may need to use this straight line with unknown slope to solve the problem, so you need to use the above writing method.

2.ac & lt0, bc<0 means that A, B and C cannot be 0, so the straight line is written in the form of oblique specification.

Y=(-a/b)(x+c/a) It can be seen that the equation passes through the fixed point (-c/a, 0), and on the right side of the X axis, the slope is >; 0, so you can see that the straight line does not pass through the second quadrant when drawing.

3. The answer to the third question is-1/3.

This is an exercise of the formula of the distance from a point to a straight line, and the answer is your answer.

This question applies to the application of your first question.

Let the equation of a straight line be y-2=k(x+ 1) and get y=kx+k+2.

Then the problem is transformed into the problem that the distance from the origin to the straight line y=kx+k+2. Bring in the distance formula from point to straight line and calculate k=- 1 or k=-7.

6. The P coordinate (0, 1), and the P point is not on the given straight line, so

Let the straight line be 2x+ 1 1y+C=0.

Then the distance from point P to 2x+11y+16 = 0 is equal to the distance from the expected straight line 2x+ 1 1y+C=0.

Introduction formula of distance from point to straight line

Get C= 16 (because it coincides with a given straight line) or C=-38.

The linear equation is 2x+ 1 1y-38=0.