A straight line is drawn through the point (1: 6), parallel to the straight line y = -6x. Find the ordinate of the point

A straight line is drawn through the point (1: 6), parallel to the straight line y = -6x. Find the ordinate of the point of intersection of this straight line with the Oy axis.

Let us write the equation of the required straight line in general form: y = kx + b.

In the initial data for this task, it is reported that the desired straight line must be parallel to the straight line y = -6x.

Therefore, the slopes of these straight lines must coincide, that is, the coefficient k must be equal to -6 and the equation of the desired straight line must be as follows: y = -6x + b.

According to the condition of the problem, this straight line passes through the point (1; 6), therefore, the following relation must be fulfilled:

6 = -6 * 1 + b,

whence follows:

b = 6 + 6 = 12.

Therefore, the desired straight line is given by the equation y = -6x + 12.

Find the ordinate of the point of intersection of this straight line with the Oy axis.

Substituting the value x = 0 into the equation of the straight line, we get:

y = -6 * 0 + 12 = 12.

Answer: 12.