Answers to the fourth and fifth questions on page 109 of the second textbook of compulsory mathematics for senior one.

4. Proof: Let the straight line l 1 intersect with l2 at point P (x0, y0).

According to the meaning of the question, a1x0+b1y0+c1= 0, A2X0+B2X0+C2 = 0.

Substituting the coordinates (x0, y0) of the target P into the left side of the equation A1X+B1Y+C1+λ (A2x+B2y+C2) = 0, we get

a 1x 0+b 1 y0+c 1+λ(a2 x0+B2 y0+C2)= 0+0 = 0

That is, the coordinates of the point P satisfy the equation A1X+B1Y+C1+λ (A2X+B2Y+C2) = 0.

So the point P is on the straight line represented by the equation A1X+B1Y+C1+λ (A2x+B2y+C2) = 0.

And because the equation a1x+b1y+c1+λ (a2x+b2y+C2) = 0 can be arranged as (a1+λ a2) x+(b1+λ b2) y.

This is a binary linear equation about x and y, which represents a straight line, so the equation

A1x+b1y+c1+λ (a2x+b2y+C2) = 0 (λ belongs to r) indicates a straight line passing through the intersection of l 1 and l2.

5.( 1) Simultaneous two linear equations, and find the intersection coordinates (-2,2).

This line is perpendicular to the known line.

So let 2x+3y+λ=0 replace the intersection coordinates.

Get 2x+3y-2=0.

(2) Simultaneous two linear equations to obtain intersection coordinates (3, 2).

This line is parallel to the known line.

So let 4x-3y+λ=0 substitute into the intersection coordinates.

Get 4x-3y-6=0.

I'm exhausted. I hope you are satisfied.