# Equate the straight line through points C (6; 2) and D (-1; -3).

The straight line is given by the equation y = kx + b. Let’s substitute the coordinates of these points into this equation.

1) C (6; 2); x = 6; y = 2; we get the equation 2 = k * 6 + b; 6k + b = 2.

2) D (-1; -3); x = -1; y = -3; we get the equation -3 = k * (-1) + b; -k + b = -3.

We combine the obtained equations into a system and, having solved it, we find the values of k and b.

6k + b = 2; -k + b = -3 – express the variable b from the second equation;

b = k – 3 – substitute in the first equation;

6k + (k – 3) = 2;

6k + k – 3 = 2;

7k = 2 + 3;

7k = 5;

k = 5/7.

Find b:

b = k – 3 = 5/7 – 3 = -2 2/7.

Substitute the found values of k and b into the equation of the straight line y = kx + b.

y = 5/7 x – 2 2/7.

Answer. y = 5/7 x – 2 2/7.

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