Lesson video

Hello, Mr. Robson here, welcome to Maths.

Today, we're doing some advanced problem-solving with linear graphs.

I love problem-solving, I know that you love problem-solving, so let's not hang around here, let's get on with it.

Our learning outcome is I'll be able to use our knowledge of linear graphs to solve problems. Some keywords we're gonna hear today, tangent and intersect.

The tangent of a circle is a line that intersects the circle exactly once.

Two parts to our learning today.

The first part is quite delightful, tangent and radius problems. On this Cartesian coordinate grid, we see a circle and a tangent to that circle.

If we draw a radius from the point where the tangent touches the circle, what do you notice about the angle created? Well spotted, it's a right angle, it's 90 degrees.

This tangent and radius meet at a right angle.

We know this, because the horizontal line y = 5 meets a segment of the vertical line x = 4, and we know that horizontal lines and vertical lines are perpendicular to each other.

If they meet, they'll meet at a right angle.

When we discover something like that in mathematics, a tangent and a radius meeting at a right angle, we want to ask the question, was that true just for that case, or is it true for all cases? Let's look at this case then.

Do this tangent and radius also meet at a right angle? And if they do, how might we show it? Pause, and have a think about those two questions.

Welcome back, they absolutely do, that's a right angle for sure, but it's not good enough to say, "It looks like a right angle," and it's not good enough to say, "I measured it with my protractor," no matter how accurate you were with your protractor.

When asked to show why something is true, we need to fully justify our reasoning.

so this tangent and radius definitely do meet at a right angle.

How do we show this is true? How do we reason and justify that? You know how to describe the equation of the line of the tangent and you know how to describe that of the line segment that is the radius, let's label these L1 and L2.

In the case of L1, that line segment is part of the line, y = x - 1, that would be a gradient of 1, and a Y-intercept of 0, -1.

for L2, that tangent is the line y = 9 - x, a Y-intercept, at 0, 9, and a gradient of negative 1.

From here, that word gradient is the key.

What do you notice about the gradients of these two lines? Well spotted, the first gradient or the gradient of the first line is positive 1.

The gradient at the second line is negative 1, you noticed, didn't you? They are negative reciprocals of each other.

Lines are perpendicular if the product of their gradients is negative 1.

positive 1 multiplied by negative 1 gives us negative 1, that means these two lines are perpendicular, that tangent and radius must meet at a right angle.

Quick check, to see if you've got this so far.

True or false? When asked to show if this tangent is at a right angle to the radius, we can just measure it with our protractor as proof.

Is that true, or is it false? Once you've decided, select one of the two statements at the bottom of the screen to justify your answer.

Pause, and do this now.

Welcome back, I do hope you said false, and justified it with a statement, to show, means we have to fully justify our reasoning.

Well done.

Will the tangent and radius always meets a right angle, even when the lines are not vertical, horizontal, or perfectly diagonal? Yes, they will, and you can see that they do in this case.

How can we see that? Well, look at the tangent, it goes 2 right 1 up, that's perpendicular to our radius, which is 2 down 1 right.

I hope you can see that it's just a 90-degree rotation of that journey, this must be a right angle, our tangent and radius are meeting at a right angle.

However, to show this is true, we need to find the equations of the lines.

Let's label them then, L1 and L2.

Line one, The tangent is y = 1/2` x + 6.

That's a Y-intercept of 0, 6, and a gradient of a half.

The radius is a segment of the line y = -2x - 4, that's a gradient of negative 2, and that line would go on to intercept the Y axis at 0, -4.

The gradients are the key.

The gradient of line 1 is a half.

The gradient of line 2 is negative 3.

Are they negative reciprocals of each other? Absolutely, they are.

A half multiplied by negative 2 is negative 1.

Therefore, these gradients are perpendicular, the lines are perpendicular, that tangent and radius are meeting at a right angle.

Have you seen that notation before, therefore, and perpendicular? We mathematicians are always seeking efficiency and we can gain some efficiency in our notations here, by using that shorthand notation.

Quick check, you can do a problem just like that one.

line segment L1 is a radius of the circle and has the equation y = -1/4 (x +17).

L1 and L2 meet at the point (7, -6).

The point (9, 2) is also on L2.

Show that L1 and L2 are perpendicular.

Over to you now, pause, and give this problem a go.

Welcome back.

Let's see how we got on, shall we? Let's look at line 2, see if we can find the equation of the line.

We'll start with finding the gradient.

As a changing y of 8 and a change in x of 2 between those two coordinates that we can see on the line, therefore, we've got a gradient of 4.

L2 must be y = 4x + c.

We could go on to find c, there'll be moments where we need to know the Y-intercept, but in this case, we only need to know the gradient because we already know that the two lines have a common point at (7, -6).

Let's compare the gradient to the two lines.

The gradient of the radius is - 1 1/4, the gradient of our tangent is 4, <v ->1/4 x 4 is - 1,</v> therefore, they are perpendicular, problem solved.

Practise time, now.

For Question 1, I'd like you to show that the line L2 is tangent to the radius; line segment L1.

My top tip, you have the scale on the axes.

You can choose whichever coordinates you'd like to help you here.

Pause, give this problem a go now.

Question 2, AB is a radius of the circle.

AB and BC meets at common point (-4, -9).

I'd like you to show that BC is tangent to the circle.

Pause, and give this problem a go now.

For Question 3.

AB is a radius of the circle.

AB and BC meets at common point (-8, 8) Izzy says, "I have measured the angle with my protractor and it is 90 degrees.

BBC is a tangent." Well, it looks like one, and it probably measures like one with your protractor, but what I'd like you to do is show Izzy why BC is not a tangent.

You've got the skills to do that, pause, and use them now.

Feedback time now, let's see how we go on.

Question 1, we were showing that line L2 is tangent to the radius; line segment L1.

You needed to choose some coordinates to help you with this problem, I went for coordinate (2, 0), (3, 3), and (6, 2).

Looking at L1, the line segment that is our radius, the gradient, we've got a change in y, 3, a change in x of 1, it's a gradient of 3.

That means line segment L1 must be on the line, y = 3x + c, we don't need to find c, we're just trying to prove that these two lines are perpendicular, so we're interested more in the gradient.

Turn our attention now to the tangent line L2.

The gradient there, between my coordinates (3, 3) and (6, 2), I've got a change in y of negative 1, a change in x of positive 3, that gives me a gradient of - 1/3.

That means the tangent must be the line y = -1/3 x + c Again, not necessarily interested in that Y-intercept, it's the gradient that interests us at the moment.

when we look at the gradients - 1/3 X positive 3, is indeed negative 1, two gradients multiply to make negative 1, they are negative reciprocals of each other, therefore, these two lines are perpendicular.

If they share that common poin (3, 3,) then our radius and our tangent, are indeed meeting at a right angle.

Question 2, we need that AB is a radius of the circle.

AB and BC meet at a common point, (-4, -9), and we're asked to show that BC is a tangent to the circle.

Let's look at the gradient of AB.

We've got a change in y of negative 3, a change in x of positive 2, that's a gradient of negative 2.

The gradient of BC, a change in y, positive 9, a change in x of positive 18, that's a gradient of 1/2.

Negative 2 multiplied by half is negative 1, therefore, they are perpendicular.

BC is a tangent because we know that tangents and radius meets as a right angle.

For Question 3, I've absolutely no doubt it looks like a 90-degree angle when you measured the Izzy, but what we were doing was showing Izzy why BC is not a tangent.

The gradient of AB is -3/2.

The gradient of BC is 5 over 8.

When we multiply the two gradients together, we do not get negative 1.

The gradients are not negative reciprocals of each other, therefore, they're not perpendicular.

They're not perpendicular, AB, BC are not meeting at right angles, that means BC is not a tangent.

Well, how nice of you to say, Izzy, you're welcome.

Onto part 2 of the lesson, where we're gonna look at some quite delightful ratio problems. We can divide a line segment on our Cartesian coordinate grid in a given ratio.

For example, find point P when AP to PB is in the ratio of 1 to 3.

There are so many ways we could tackle this problem.

I'm gonna consider the horizontal journey and the vertical journey from A to B.

We're travelling 8 in the x direction and 4 in the y direction.

Let's divide those journeys by our given ratio of 1 to 3.

The horizontal journey will become 2 to 6, and our vertical journey will become 1 to 3.

What do I mean by that? I'm gonna turn that journey into those two journeys.

The horizontal, 8 being divided into a ratio of 2 to 6, and the vertical being divided into a ratio of 1 to 3.

They're two similar triangles with a scale factor of 3, funnily enough, that means point P is there.

Point P is at the coordinate (-6, 1).

We can use this skill to solve more complex problems. Point P is such that AP to PB is in the ratio 1 to 3.

Find the area of triangle APO.

The first bit we already did.

We already know point P, it's there.

The second bit is new.

Find the area of triangle APO.

Ah, there's the triangle APO, I need to find the area.

In this case, we've got a base of 8, a vertical height, or an altitude of 1, and we use the formula for every triangle it's base X height over 2, so we do 8 multiplied by 1 over 2, and we get an answer of 4 square units.

Quick check, you can do that.

Point P is such that AP to PB is in the ratio of 3 to 1.

Find the area of triangle APO.

Pause, and give this problem a go.

Welcome back, let's see how we did.

The first step was to find point P.

This time, I turn that ratio around, instead of 1 to 3, it reads 3 to 1, that means P is there this time.

Next up, we're finding the area of triangle APO.

That's triangle APO.

You should have noticed a base of 8 and a vertical height, or altitude of 3, so we're doing 8 by 3 over 2, and we should have got the answer, 12 square units.

There are lots of complex problems of this variety that we might encounter.

Points A, P, and B are on the line 3y - x = 15.

Point P is such that AP to PB is in the ratio of 3 to 2.

Find the equation of the line that goes through points P and O.

Wow, there's an awful lot going on there, a lot for me to digest.

I think I'm gonna take it step by step.

Let's look at this first thing that we're told.

Points A, P, and B are on the line 3y - x = 15.

Ah, I can work with that.

When x equals 0, y equals 5.

That tells us that coordinate (0, 5) is on this line, oh look, it's coordinate B.

Alternately, when y is 0, the x value has to be negative 15.

So we've got a coordinate of (-15, 0 ) on this line.

Aha, we found coordinate A.

Once we've used that first bit of information, we'll look at the second bit of information, which was, point P is such that AP to PB in the ratio of 3 to 2.

There's a lot of ways we can solve this problem.

To get from A to B, we need to travel 15 in the positive x direction and 5 in the positive y direction, but we don't wanna get from A to B because we're dividing the line in the ratio of 3 to 2.

The ratio of 3 to 2 means we've got 5 parts, but I only want to travel 3 of those parts.

I'm gonna do 3/5 of that journey.

3/5 of 15 is 9, 3/5 of 5, the y journey, that's 3.

So point P must be along 9 and up 3 from point A, in that case, point P is the coordinate (-6, 3).

What was next in our problem? Oh yes, the problem we were asked to solve.

We were asked to find the equation of the line that goes through points P and O.

Well, we now know point P is -6, 3.

We can draw a line through P and O, and we know the gradient of that line.

To get from P to O, we'd go negative 3 in the y direction, and positive 6 in the x direction, that'll give us a gradient of negative a half.

What else do we know about the line? The Y-intercept is at the origin, so this must be the line y = - 1/2 x.

What a lovely problem.

An awful lot going on there, but when we break it down into its little component parts, we find it's all very digestible.

Quick check, if you can do a problem like that.

Points A, P, and B are on the line 3y + x = 30.

Point P is such that AP to PB is in the ratio of 3 to 2.

Find the equation of the line that goes through points O and P.

Remember, go slowly, take it step by step, give it a good go.

Pause, do that now.

Welcome back, let's see how we did.

Hopefully, we noticed from the equation 3y + x = 30, when x equals 0, y equals 10, that must be coordinate A.

when y equals 0 x equals 30, that must be coordinate B.

Point P is such that AP to PB is in the ratio of 3 to 2, so two similar triangles in the ratio of 3 to 2, means point P must be there at (18, 4).

Next step, we want to know this line that goes through O and P, let's find the gradient.

A change in y of 4, a change in x of 18, the gradient must be 2 over 9.

It goes through the origin, so this is the line y = 2/9 of x, problem solved.

Practise time now, Question 1, points A, P, and B are on the line y - 2x +10 = 0 Point P is such that AP to PB is in the ratio of 4 to 1.

For part A, I'd like you to find the area of triangle OBP, and for part B, I'd like you to find the area of triangle OPA.

Pause, give these problems a go now.

Question 2, point A has coordinates (-4, 1), and point P has coordinates (0, 2), and point O is the origin.

Point P is such that AP to AB is in the ratio of 2 to 5.

Find the area of triangle OPB.

My hint is that a visual representation may be useful here.

You don't absolutely have to have one, but if I were doing this problem, I'd be drawing something to help me with my thinking.

Pause, give it a go now.

Welcome back, feedback time.

Let's see how we did.

Problem 1, points A, P, and B are on the line y - 2x + 10 = 0.

When x equals 0, y equals negative 10, that's coordinate A, (0, -10).

When y equals 0, x equals 5, that's coordinate B, (5, 0).

If point P is such that AP to PB is in the ratio 4 to 1, we could turn the journey AB into 5 steps.

Why 5 steps? Because the ratio of 4 to 1 has 5 parts, so when we divide that line into the ratio of 4 to 1, P must be there.

That would be the coordinate (4, -2).

Once we know that, we're ready to tackle Part A and B.

Find the area of triangle OBP.

That's triangle OBP.

When we substitute into the formula, we could call the base 5, the vertical height, or the altitude 2, therefore, that is an area of 5 by 2 over 2, that's 5 square units.

For the area of triangle OPA, when substituting in, we could call that a base of 10, a vertical height of 4, or an altitude of 4, so the area of that triangle is 10 by 4 over 2, that's 20.

Reassuringly, for this problem, like many problems in maths, there's multiple ways to solve it.

Alternatively, you might have looked at the area of triangle AOB.

Multiply 5 by 10 divided by 2, area of AOB is 25.

The ratio can be applied to the area since one length, AB has been divided this way.

If the area of triangle OPA, so the area of triangle OBP is in the ratio of 4 to 1, we'll divide 25 in the ratio of 4 to 1 and we'll get 20 and 5.

Then look, the area of triangle OPA is 20.

The area of triangle OBP is 5.

It's the exact same result using a different method.

If you can do the same problem using two different methods and you arrive at the same result, that's reassuring, you kind of know you're right.

Question 2.

We were given a lot of information and my recommendation is that you turn this into a visual representation.

There, I can see A has coordinates (-4, 1), P has the coordinates (0, 2), point O is the origin.

Then we were told that point P is such that AP to AB is in the ratio of 2 to 5, so I need to scale this journey up from 2 to 5.

The horizontal journey will go from 4 to 10, and the vertical journey will go from 1 to 2.

5, so to get from A to B, I need to do that journey 10 in the positive x direction, 2.

5 in the positive y direction.

That means B is there at the coordinate (6, 3.

5).

We were then asked to find the area of triangle OPB.

There, it will be that triangle.

When substituting into the formula for the area of a triangle, we'll call the base 2, the vertical height or the altitude 6, and we'll get an area of 2 by 6 over 2, that's 6 square units.

I enjoyed that problem, that was nice! Sadly, we're at the end of the lesson now.

We've learned that we can use our knowledge of linear graphs to solve a variety of problems in mathematics.

For example, we were able to show that a tangent and radius of a circle meet at a right angle with our knowledge of the gradients of perpendicular lines.

I hope you enjoyed this lesson as much as I did, and I look forward to seeing you again soon for more lovely mathematics.

Goodbye, for now!.