### Video Transcript

Find the equation of the straight line that is parallel to the π¦-axis and passes through the point of intersection of the two straight lines π¦ equals negative three and π₯ equals eleven fifteenths π¦.

Sketching in an π₯π¦-plane, where the axes cross at our origin, we can draw in the two straight lines given to us as π¦ equals negative three and π₯ equals eleven fifteenths π¦. π¦ equals negative three is a horizontal line that crosses the π¦-axis at negative three, while π₯ equals eleven fifteenths π¦ passes through the origin and has a positive slope like this. The equation of the straight line we want to solve for is parallel to the π¦-axis and it passes through the point of intersection of these two lines.

A line thatβs parallel to the π¦-axis is one that has a constant π₯-value. In other words, it would be a vertical line. And we know it passes through our point of intersection. We can completely describe this straight line by writing out the π₯-value of all points on the line. That π₯-value will be the π₯-coordinate of this point of intersection of our two orange lines. Regarding that point, we know its π¦-value is negative three because the point lies along the line π¦ equals negative three.

To find the corresponding π₯-value, we can use the fact that this point of intersection also lies along the line π₯ equals eleven fifteenths π¦. At the intersection point, we know that π¦ equals negative three. So if we substitute that in for π¦ in this equation, then π₯, we see, equals negative eleven-fifths. The coordinates of our point of intersection then are negative eleven-fifths, negative three. And therefore, this line consists simply of every point in the π₯π¦-plane, where π₯ equals negative eleven-fifths. That fact lets us write the equation of this straight line. The equation of the line parallel to the π¦-axis and passing through the given point of intersection is π₯ equals negative eleven-fifths.