Equate the straight line through points C (6; 2) and D (-1; -3).
May 28, 2021 | education
| 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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