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Hello everyone.

I hope you're all doing well.

Welcome to another maths lesson with me, Mr. Gratton.

Today we'll be looking at constructing a perpendicular to align through any point using the properties of a Rhombus.

Pause here to check some of the definitions of the keywords we'll be using throughout today's lesson.

We'll first look at constructing a perpendicular through a point where we need to shorten the length of the line segment to do so effectively.

Here's a quick comparison of some of the key points you may have come across previously.

A perpendicular bisector to AB is a line that intersects the midpoint of AB at a right angle.

This divides the line segment AB into two equal parts.

In this example, the original line segment was 20 centimetres long and the perpendicular bisector divides it into two equally long 10 centimetre parts.

In comparison, a perpendicular not bisector to AB can be any line that intersects at a right angle at any point on the line segment AB.

That 20 centimetre long line segment can be divided into two unequal parts such as this 3 centimetre, 17 centimetre split.

Jacob is convinced that the construction of a perpendicular is only possible if it is also a bisector because he's only familiar with the construction method for a perpendicular bisector.

However, Laura is convinced that the line segment can be shortened so the point becomes the new midpoint of the modified line segment.

Whilst Jacob agrees, he wonders how.

Laura is correct.

It is possible to construct a perpendicular for any point on line segment AB.

The goal is to create a shortened line segment which has the point as the centre of a circle and therefore the midpoint of that shortened line segment like so.

To do this, place the compass needle on the point where you want to construct a perpendicular through.

Set your compass width to any length where the pencil end touches the line segment on either side of the line, like this.

Notice how the pencil end of these pair of compasses still touches AB both in the direction of A and in the direction of B.

When you're satisfied that your compass width is appropriate, draw a full circle.

The parts of AB that are outside of the circle can now be ignored.

They will not be used in the construction of this perpendicular.

The process now becomes identical to constructing a perpendicular bisector.

Now with the two points that the circle intersects AB as the new endpoints of the shortened line segment.

Place the compass needle on one intersection of the circle under AB and change the width of the compass to greater than half the length of the diameter of the circle.

When you're satisfied that your compass width is appropriate, draw two arcs, like so.

By ensuring the compass width is the same, repeat on the other intersection of the circle and AB, like so.

Note the arcs can intersect anywhere, inside, outside, or on the circle itself.

This makes no difference to the construction.

Once you have your two intersections between the four arcs, draw a straight line through these two intersections to complete the perpendicular through that point.

Like so.

Whilst we've seen a successful method, Laura wonders what happens if the circle drawn is too big, like this.

Jacob notes that the circle will not intersect AB twice, and so the point will not be the midpoint of this modified line segment.

The two parts of the modified line segment will still be of different lengths, and this actually means the circle will have no purpose.

As Laura notes, we want both the left and the right parts of this modified line segment to be the same length, so we can construct a perpendicular bisector through that point.

If we were to construct the perpendicular bisector of this modified line segment, the part of the line segment inside the circle, the bisecting line would not pass through the point that we wanted it to.

For this demonstration, I will construct the perpendicular with the description on the left.

After each set of steps that I go through, perform the same set of steps for the construction of a perpendicular to a nine centimetre line segment With the details on the right.

Using a ruler, I will draw an eight centimetre line segment and mark on three centimetres and label that point, "C" And label the endpoints of that line segment, "AB".

Pause now to do the same with a nine centimetre line segment with a four centimetre point marked on.

Pause now to do this.

Now that we have our line segment, place the compass needle on point C and set the compass width to any length where the pencil end touches the line segment on both sides of point C.

And when you are satisfied with your compass width, draw a full circle.

Pause now to draw a circle with centre C that intersects with AB twice.

Do not worry if your circle ended up being too big.

You can just draw another smaller one with a centre still at point C.

Once you've drawn your circle, place the compass needle on one intersection between the circle and the shortened AB and extend the pencil end beyond C so that the compass width is over half the diameter of that circle.

After you've set your compass width appropriately, draw two arcs one on either side of that line segment like so.

Pause now to draw two arcs with the centre at one of the intersections of the circle and line AB.

Try this yourself.

By keeping the compass width the same as the previous two arcs, construct two more arcs with the compass needle on the other end of the shortened line segment where the circle intersects the original line segment like so.

Pause now to do this for your construction.

Now that you have your two intersections of those four arcs, draw a line through both of those intersections.

This is the perpendicular to AB through point C, like this.

Pause now to try this for yourself to get this perpendicular bisector.

When you've drawn your perpendicular bisector, use a protractor to measure that angle between your line segment and the perpendicular to see how close to 90 degrees that angle is.

Here's a quick check from Andeep.

Andeep draws a circle with a centre at point C using a pair of compasses whose width has not changed.

Find the distances A and B.

Pause now to think about the relationship between the compass width and the radius of the circle.

Both A and B have a length of five centimetres.

This is because the compass width is identical to the radius of the circle that you draw with that pair of compasses.

Next check from Sophia.

Sophia has incorrectly constructed a perpendicular for AB through point C.

What did Sophia do wrong? Pause now to look through all of these options.

The circle is simply too big and so it does not intersect with AB twice.

Sophia has not drawn all of the expected construction arcs.

There should be a total of four for a total of two intersections between these four arcs.

As a result, the line through C may not be a perpendicular to AB.

And lastly, Alex has partially completed a construction for a perpendicular of AB through the point E.

What is incorrect about Alex's pair of compasses? Pause now to consider which option is correct for Alex's construction.

And the answer is C.

The compass width is too short.

When constructing the arcs, the width of the compass should always be over half the diameter of the circle, and this means that the pencil end should be beyond point E.

Okay, onto the independent practise.

For question number one, complete the construction of the perpendicular through point Z of each of these five line segments.

Pause now to do all five.

And question number two is asking you to construct two different perpendiculars, one through line segment AB and one through line segment BC.

Draw each perpendicular long enough so that they intersect each other.

Label the point that the two perpendicular intersect point Z and then attempt to identify the quadrilateral bound by the vertices XBY and the point that you identified Z and justify why you think the construction creates that shape.

Pause now to do question two.

And following on from question number two, come up with another two perpendicular constructions that create a different quadrilateral.

Using a protractor, measure each of the four interior angles of the quadrilateral that you have constructed.

Explain how you know what type of quadrilateral you have constructed.

Pause now to start your creation.

And onto the answers.

Here are all the constructions for question number one.

Pause here to check that your constructions match the ones that are on screen.

And for question number two, here are where your perpendiculars should have gone, and the quadrilateral that would've been created was a kite.

This is because the line segments XB and BY are of the same length, and XZ and YZ are of the same length.

Furthermore, the angle ZXB is equal to the angle ZYB.

And for question number three, many different quadrilaterals could have been constructed such as a rhombus or a parallelogram.

We've looked at perpendiculars where we modify the length of a line segment by shortening the line segment, but there may also be situations where extending the line segment is necessary.

Let's have a look why.

Jacob is convinced that he spotted a point where the perpendicular cannot be constructed.

Perhaps the point is at the end point of a line segment, but Laura questions whether it's impossible or not.

Jacob's reasoning is that any circle must intersect the line segment exactly twice, and at an endpoint, the circle is guaranteed to intersect only once.

To which an answer is given.

Can't you just extend the line segment until it intersects the circle twice like so? Laura is correct.

You can extend the length of a line segment in order to make it intersect with the circle twice.

Here's another demonstration.

This one to show that Laura is correct.

It is possible to construct a perpendicular to the endpoint AB.

Pause here to draw a line segment AB of five centimetres and mark the endpoint A.

Try this yourself.

Place the compass needle at point A and draw a circle.

Pause the video here to set your compass width to any suitable length.

Place your compass needle on point A and draw your circle, like this.

Place your ruler carefully over AB.

Extend the line segment until it intersects the circle a second time.

Pause now to give yourself time to extend your line segment.

And now our method is identical to a perpendicular bisector.

Place the compass needle at one intersection between the extended line segment and the circle, and make a width of the compass needle over half the diameter of the circle and draw two arcs.

Pause now to draw two suitable blocks of your own.

And by making sure that the compass width is the same as the previous two arcs, draw another two arcs with the compass needle at the other intersection between the circle and the line segment.

Pause now to do this yourself.

As usual, the perpendicular is a line that passes through both these intersections.

Pause now to complete your construction.

Whilst our demonstration constructed a perpendicular through the endpoint of a line segment, extending is also useful when the point that the perpendicular needs to pass through is close to an endpoint.

Without extending this line segment, the circle is far too small to be used effectively.

It is more important that you choose a circle that is a suitable length for the pair of compasses that you have and commit to extending the line segment rather than forcing a circle that is too small for the sole reason of not needing to extend the line segment.

Whose partially completed construction of a perpendicular that requires an extended line segment is correct? Pause now to look through all four constructions.

The correct answer is Jacob.

Jacobs' is the only construction that passes through the circle twice where the line segment has not been extended at a random angle like Sophia's.

And here's question one of the next practise task.

Construct a perpendicular to each line segment through the point X and then use a protractor to measure how close your construction is to a 90 degree perpendicular line.

Pause now to do this.

And for question number two, plot the points A, B, C, and D on this coordinate grid and then find the perpendiculars to pairs of points AB, AC, and BD.

Pause now to do question two.

And here are the answers.

For question number one, here are where your perpendicular lines should have gone.

And for question number two, here are all the line segments and the perpendiculars that you should have drawn.

And for part D, the perpendiculars to AC and BD are parallel because the line segments AC and BD themselves are also parallel to each other.

So we've looked at constructing perpendiculars to midpoint on a line segment, to points anywhere on a line segment and to a point at the end point on a line segment.

But is it possible to construct a perpendicular to a line segment that passes through a point off of the line segment? Let's take a look.

Laura thinks that no matter where a point is, the initial aim is to draw a circle that intersects the line segment twice, even if the point isn't on the line segment itself.

Laura thinks the method will not change.

A circle that intersects the line segment twice is still the goal, even if you need to extend the line segment to achieve this.

Now that we know that all we need to do to make a perpendicular construction possible is to have a line segment that is intersected by a circle twice where the centre of that circle is the point that the perpendicular needs to pass through, we can therefore construct the perpendicular to a line segment through a point that is anywhere, on or off the line segment.

If the point is off of the line segment, it is more likely that we have to draw a bigger circle so that it can reach and intercept the line segment and extend the line segment so that it intersects with the circle twice.

Onto the next few checks.

Which of these circles will help in the construction of the perpendicular to QP through the point Z? Pause to look carefully at the three circles and think what can be done so that it intersects with QP twice.

The two answers are B and C.

Extending QP will allow it to intersect with either B or C.

However, A will never intersect with the line segment QP.

Sometimes finding the perpendicular through a point off of a line segment can help find the shortest distance from that point to the line segment.

However, this is only true if the perpendicular intersects part of the original line segment and not an extension of it.

For example, find the shortest distance from C to the line segment AB.

Draw a circle with centre C.

AB needs to be extended so it intersects with the circle twice.

As the perpendicular intersects the original line segment of AB, we can measure that the shortest distance from C to AB is 4.

3 centimetres.

However, for point D, after drawing a circle extending the line segment and constructing the perpendicular, we can see that the perpendicular intersex, the extension of the line segment AB not its original length.

Therefore, 4.

3 centimetres is not the shortest distance to AB, only the shortest distance to an extension of the line segment AB.

The true shortest distance is 5.

1 centimetres, which is the distance from D to one of the endpoints of the original line segment, in this case, the endpoint B.

Okay, onto the final check.

Complete each statement with a value from the diagram.

Pause now to look through the diagram and think about what each length represents.

Seven centimetres is the shortest distance to the extension of SR.

Whilst 12 centimetres is the shortest distance to the actual original line segment SR.

Whilst 10 centimetres is the radius of the circle.

Okay, final set of practise questions.

For question one, construct five perpendiculars to AB through each of the five points.

And for part B, measure the shortest distance from each point to the line segment AB itself.

Pause now to to give all five of these points a go.

And the final question.

A boat out at sea needs to find the closest port so it can refuel.

Which port is the boat closest to, and at which part of the port should the boat dock so it travels the shortest distance? Pause now to do this final question.

And for the answers to question number one, here are the five perpendiculars.

Note that X is the only point whose perpendicular intersects the extension of AB rather than the original line segment AB itself.

And for question number two, port B is the closest.

You have to be careful though because the shortest distance to an extension of port C is closer, but the actual port C is further away.

Thank you so much for all your effort today in a very hands-on constructions lesson where we have constructed perpendiculars to a point on a line segment by either shortening or extending the length of the line segment where the perpendiculars to a line segment can also be constructed through a point off of that line segment.

Regardless of where that point is, you can always draw a circle with its centre at that point and a radius of a circle large enough for the circle to intersect that line segment or extension of it twice.

We've also considered that if a perpendicular intersects the original line segment, the distance from the point to the line segment on the perpendicular pathway is the shortest distance.

Thank you so much for joining me today.

And until next lesson, take care and have an amazing rest of your day.