1 Let the equation of this straight line y- 1=k(x-2), and the straight line 5X-2Y+3=0 is y=5x/2+3/2, indicating that the slope of this straight line is 5/2.

With tan 45 = | (k-5/2)/(1+5k/2) | =1,we can get k=3/7 or k=-7/3.

So this straight line is y- 1 = 3/7 (x-2) and y- 1 =-7/3 (x-2), which means 3x-7y+ 1=0 or 7x+3y- 17 =

2.( 1) Because two straight lines are parallel, the slopes of the two straight lines are equal, that is, the slopes of the straight lines are equal to 5. Therefore, Y=5X is obtained.

(2) Let the linear equation y+2=k(x- 1) and the linear equation X-2Y+ 1=0 be 1/2. According to the meaning of the question, k*( 1/2)=- 1.

So this straight line is y+2=-2(x- 1), which means 2x+y=0.

The main applications of these two problems (the slopes of two straight lines are k 1, k2 respectively) are as follows.

The included angle formula tana = | (k1-k2)/(1+k1k2) |

Two straight lines are parallel, k 1=k2.

Two straight lines are perpendicular to k 1*k2=- 1.