Given a triangle ABC, find the equation of a straight line passing through point A, parallel to BC

Given a triangle ABC, find the equation of a straight line passing through point A, parallel to BC, where A (1; 3), B (3; 5) and C (7; -1)

Let’s write the equation of the straight line that passes through the points B (3; 5) and C (7; -1):

(x – 3) / (7 – 3) = (y – 5) / (-1 – 5).

Simplifying the written equation, we get:

(x – 3) / 4 = (y – 5) / (-6);

(-6) * (x – 3) = 4 * (y – 5);

(-6) * (x – 3), / 2 = 4 * (y – 5) / 2;

(-3) * (x – 3) = 2 * (y – 5);

-3x + 9 = 2y – 5;

3x + 2y – 5 – 9 = 0;

3x + 2y – 14 = 0.

Any straight line parallel to a given one can be written using the equation 3x + 2y + c = 0, where c is some number.

Let us find at what value of the parameter c the straight line described by this equation passes through the point A (1; 3):

3 * 1 + 2 * 3 + c = 0;

9 + s = 0:

c = -9.

Therefore, the equation of the straight line passing through point A parallel to the side BC is 3x + 2y – 9 = 0.

Answer: the desired equation of the straight line is 3x + 2y – 9 = 0.