### Video Transcript

Find the vector equation of the straight line that passes through the point of intersection of the two straight lines negative eight π₯ minus π¦ is equal to seven and negative five π₯ minus three π¦ is equal to two and the point 12, eight.

In this question, we are asked to find the vector equation of a straight line. Weβre told that this straight line passes through two different points. First, it passes through the point of intersection between the two straight lines negative eight π₯ minus π¦ is equal to seven and negative five π₯ minus three π¦ is equal to two. Weβre also told that this line passes through the point with coordinates 12, eight.

To answer this question, letβs start by recalling what we mean by the vector equation of a line. The vector equation of a line is a line in the form π« is equal to π« sub zero plus π times π, where the vector π« sub zero is the position vector of a point on the line and the vector π is a direction vector of the line. Itβs any nonzero vector which runs parallel to the line, and π just represents any scalar value. So we need to determine the position vector of a point which lies on the line and the direction vector of this line. And weβre already told that the line passes through the point with coordinates 12, eight. And the position vector of this point is the vector 12, eight. We can set this equal to our vector π« sub zero in the vector equation of this line.

This means all we need to determine is the direction vector of the line. And we can find the direction vector of a line by using the coordinates of two points on the line. This means weβre going to need to find the coordinates of another point on the line. And weβre going to have to do this by finding the point of intersection between the two given straight lines. And we can do this by recalling to find the point of intersection between two given straight lines, we just need to solve them as simultaneous equations. In other words, the π₯- and π¦-coordinates of a point of intersection between the two lines must satisfy the equation of both lines.

And thereβs many different ways we can solve these two equations simultaneously. Weβll only go through one of these. Weβre going to rewrite the first equation to make π¦ the subject. We add π¦ to both sides of the equation and subtract seven from both sides of the equation. This gives us that π¦ is equal to negative eight π₯ minus seven. We can now substitute this expression for π¦ into the second equation. This then gives us that negative five π₯ minus three times negative eight π₯ minus seven is equal to two.

This is now an equation in π₯, so we need to solve this equation for π₯. To solve this equation, weβll start by distributing negative three over the parentheses. We get negative five π₯ plus 24π₯ plus 21 is equal to two. We can then continue simplifying this equation. Adding the π₯-terms together, we get 19π₯. And then we can subtract 21 from both sides of the equation to get 19π₯ is equal to negative 19. We can then divide both sides of this equation through by 19 to see that π₯ is equal to negative one. This is the π₯-coordinate of the point of intersection between the two lines. We can substitute π₯ is negative one into the equation of either of these straight lines to determine the π¦-coordinate of the point of intersection.

Weβll substitute this into the equation of the first line. And itβs worth pointing out this is equivalent to substituting into the rearrangement of this equation. In either case, we can solve the equation to show that our value of π¦ is one. We now have the coordinates of two points which lie on this line. This will allow us to calculate the direction vector of this line. We can do this by recalling the direction vector of a line passing through two distinct points π₯ sub one, π¦ sub one and π₯ sub two, π¦ sub two is the vector π₯ sub one minus π₯ sub two, π¦ sub one minus π¦ sub two. And itβs worth pointing out that any nonzero scalar multiple of this vector is a direction vector of the line.

Weβre told in the question our line passes through the point with coordinates 12, eight. And weβve just found the point of intersection between the two lines. This is the point negative one, one. Since our line passes through this point, we can set π₯ sub one, π¦ sub one to be 12, eight and π₯ sub two, π¦ sub two to be negative one, one. This will allow us to find a direction vector of this line. We get π is equal to the vector 12 minus negative one, eight minus one. And we can evaluate each of these components separately. These are the vector 13, seven.

Now all thatβs left to do is substitute the position vector and direction vector into the vector equation of the straight line. We get that π« is equal to the vector 12, eight plus π times the vector 13, seven, which is our final answer.