Skip to 0 minutes and 12 secondsHello. Welcome to the section of exercises of the week three. Remember to try to solve yourself the exercises before watching the videos, and before looking at the solutions that you can find in the PDF file at the bottom of the page. OK. It's Your Turn on the geometry of the plane. We have the following exercise. We have to find the equation of the line r passing through the point of coordinates (6, 2) and forming with the y-axis an angle theta equal to pi over 6. OK. Clearly our line is not a vertical line, therefore its equation has the following shape. y equal to m1 x plus q1. m1 is the slope. q1 is the y-intercept.

Skip to 1 minute and 17 secondsBut which is the meaning of these two quantities? m1 is equal to the tangent of the angle formed by our line r with the x-axis. That is the tangent of this angle. But now, looking at the picture, we have information about the angle theta that our line forms with the y-axis. And clearly, also, this angle here is again theta. Therefore the angle we are interested in between our line r and the x-axis is equal to what? To pi over 2 minus theta. Then the slope m1 is equal to the tangent of pi over 2 minus theta. That is, the tangent of pi over 2 minus pi over 6, which is the tangent of what?

Skip to 2 minutes and 31 secondsWe have six at the denominator, and here we have 3 pi minus pi, which is 2 pi. That is the tangent of pi over 3. By definition, the tangent of pi over 3 is equal to the sine of pi over 3 over the cosine of pi over 3. The first is the square root of 3 over 2, and the second, the cosine, is 1 over 2. And we get the square root of 3. OK. This is m1. And what we can say about q1? q1 is the y-intercept. The y-intercept is the second coordinate of the unique point belonging to our line which belongs also to the y-axis.

Skip to 3 minutes and 30 secondsThat is, q1, as you can read easily from this equation, is exactly the value of y when x is equal to 0. Good. Therefore let us compute who is this q1.

Skip to 3 minutes and 50 secondsHow can we compute q1 now. Substituting what? The coordinates of our point. You know (6, 2) are the coordinates of a point belonging to our line. Therefore y equal to 2 and x equal to 6 are solutions of this equation. And we immediately get 2 equal. OK. m1 we know is the square root of 3 times the first coordinate, 6, of our point, plus q1. And then we get that the q1 is equal to 2 minus 6 times the square root of 3. Good. And now we have our equation, the equation of our line. The line r has equation y equal to the square root of 3 times x, plus 2 minus 6 times the square root of 3. OK.

Skip to 4 minutes and 59 secondsThen the second question is to compute the intersection q of r with the y-axis. It's exactly the point we were discussing before. It's that unique point in the line r which has first coordinate equal to 0. Then it's very easy, knowing the equation of our line, to understand which are the coordinates of this point q, because q is the point which has first coordinate equals 0 and the second coordinate equal to the y-intercept of our equation-- which is exactly 2 minus 6 times the square root of 3, as one can easily obtain just setting x equals 0 in the equation of the line.

Skip to 6 minutes and 0 secondsAnd then finally, we want to compute the equation of the line s passing through the point q and perpendicular to r. Therefore we want to compute the equation of this line. It's clear that also the line s is not a vertical line. Therefore, also the line s will have an equation of the shape y equal to m2 x plus q2. OK. This time, we first compute the y-intercept, because it's very easy. Because we know by the question of the exercise that our line s has to pass through the point q. Then q2 has to be equal to the second coordinate of the point q, which is 2 minus 6 times the square root of 3. And what about the slope?

Skip to 7 minutes and 15 secondsWhen you have two perpendicular lines-- non-vertical lines, both not vertical-- then the slopes have to satisfy a precise equation. Precisely, minus 1 has to be equal to m1 times m2. OK. But we know that m1, the slope or the line r, is equal to the square root of 3.

Skip to 7 minutes and 49 secondsAnd then, from square root of 3 times m2 equal to minus 1, we get immediately m2 equal to minus 1 over the square root of 3. Then we can easily conclude that the our line s has equation y equal to minus 1 over the square root of 3 times x plus the y-intercept, which is 2 minus 6 times the square root of 3. This is the equation of our line s. OK. Thank you very much for your attention.

It's your turn on geometry of the plane: points, segments, lines

Do your best in trying to solve the following problems. In any case some of them are solved in the video and all of them are solved in the pdf file below.

Exercise 1.

Find the equation of the line \(r\) passing through the point \((6,2)\) which forms an angle \(\theta=\dfrac{\pi}6\) with the \(y\)-axis. Compute the intersection \(Q\) of \(r\) with the \(y\)-axis and the equation of the line \(s\) passing through \(Q\) perpendicular to \(r\).

Exercise 2.

Given the points \(A=(2,-3)\), \(B=(4,0)\), \(C=(-1,2)\),

  1. find the coordinates of the midpoint \(M\) of \(AB\) and of the midpoint \(N\) of \(BC\);

  2. find the centroid \(G\) of \(ABC\).

Exercise 3.

Find the point of the form \((x,3/2)\) that lies on the segment between \((-2,1)\) and \((7,3)\).

Exercise 4.

Find the values of \(p\) and \(q\) such that the lines \[ r_p: pX-2Y-1=0 \quad\text{and}\quad s_q: 6X-4Y-q=0 \]

  1. have exactly one point in common;

  2. are parallel and distinct;

  3. coincide.

Exercise 5.

In the cartesian plane consider the points \(A(-2, -1)\) and \(B(1, 2)\).

  1. Find the equation of the line \(r\) passing through \(A\) and \(B\).

  2. Find the point \(C\) of intersection between the line \(r\) and the \(y\)-axis.

  3. Find the equation of the line \(s\) parallel to \(r\) and passing through the origin.

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Advanced Precalculus: Geometry, Trigonometry and Exponentials

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