It is known that the straight line given by the equation y = kx + b passes through points A (4; -6) and B (-8; -12)

It is known that the straight line given by the equation y = kx + b passes through points A (4; -6) and B (-8; -12). Find k and b, as well as the coordinates of the point of intersection with the line 2x + y = 2.

1) Find the coefficients k and b. To do this, we substitute the coordinates of these points into the equation of the straight line y = kx + b. Let’s combine the equations into a system and solve it.

A (4; -6); x = 4; y = -6; -6 = 4k + b;

B (-8; -12); x = -8; y = -12; -12 = -8k + b.

{4k + b = -6; -8k + b = -12.

Let us express the variable b from the first equation of the system in terms of k.

b = -4k – 6.

Substitute the expression (-4k – 6) in the second equation of the system instead of b.

-8k + (-4k – 6) = -12;

-8k – 4k – 6 = -12;

-12k = -12 + 6;

-12k = -6;

k = -6: (-12);

k = 0.5;

b = -4k – 6 = -4 * 0.5 – 6 = -2 – 6 = -8.

We get the equation of the straight line y = 0.5x – 8.

2) Find the point of intersection of the straight lines y = 0.5x – 8 and 2x + y = 2. To do this, you need to combine the equations into a system and solve it.

{y = 0.5x – 8; 2x + y = 2.

In the second equation of the system, instead of y, we substitute the expression (0.5x – 8).

2x + 0.5x – 8 = 2;

2.5x = 2 + 8;

2.5x = 10;

x = 10: 2.5;

x = 4;

y = 0.5x – 8 = 0.5 * 4 – 8 = 2 – 8 = -6.

(4; -6) – coordinates of the intersection point.

Answer. k = 0.5, b = -8; (4; -6).