### Video Transcript

Find the equation of the straight line that makes an angle of three π over four measured in radians with the positive direction of the π₯-axis in standard position in the unit circle.

Letβs begin by sketching the unit circle as shown. We know that we measure any angle in standard position from the positive π₯-axis. And if this angle is positive, as in this question, we measure in a counterclockwise direction. As there are two π radians in a full circle, we can mark on π over two, π, three π over two, and two π radians. Since π over two is equal to two π over four and π is equal to four π over four, our angle of three π over four radians will lie in the second quadrant. We know that any point that lies on the unit circle has coordinates cos π, sin π. This means that in this question point π has coordinates cos three π over four, sin three π over four.

In order to find the equation of the straight line that passes through point π as shown, weβll firstly calculate the exact coordinates of point π. As already mentioned, point π lies in the second quadrant. This means that its π₯-coordinate will be negative and π¦-coordinate positive. On our diagram, we can create a right triangle with reference angle π over four. The hypotenuse has length one unit, and the other two sides have length equal to the magnitude of sin π over four and a magnitude of cos π over four.

π over four or 45 degrees is one of our special angles. And both the sin and cos of π over four are equal to root two over two. This means that the cos of three π over four is equal to negative root two over two, and the sin of three π over four is root two over two. Point π has coordinates negative root two over two, root two over two.

Our straight line also passes through the origin, which has coordinates zero, zero. And we need to find the equation of this line. We could do this using the slopeβintercept or pointβslope form of the equation of a straight line. However, if we know two points π₯ sub one, π¦ sub one and π₯ sub two, π¦ sub two that lie on our line, then the equation of the line can be given by π¦ minus π¦ sub one over π¦ sub two minus π¦ sub one is equal to π₯ minus π₯ sub one over π₯ sub two minus π₯ sub one.

If we let π₯ sub one, π¦ sub one be the origin and π₯ sub two, π¦ sub two be point π, then substituting these into our equation, we have π¦ minus zero over root two over two minus zero is equal to π₯ minus zero over negative root two over two minus zero. This simplifies to π¦ over root two over two is equal to π₯ over negative root two over two, which in turn simplifies to π¦ is equal to negative π₯. The equation of the straight line that makes an angle of three π over four with the positive direction of the π₯-axis is π¦ is equal to negative π₯.

Looking at our diagram, this makes sense, as the straight line has a gradient or slope of negative one and a π¦-intercept of zero as it passes through the origin. It is worth noting that the slope π is equal to the tan of angle π. And since tan π is equal to sin π over cos π, the tan of three π over four is equal to negative one, which confirms that the slope of our line is negative one. And the equation of the straight line is π¦ is equal to negative π₯.