Video: Pack 3 β€’ Paper 1 β€’ Question 25

Pack 3 β€’ Paper 1 β€’ Question 25


Video Transcript

𝑃 is the point negative three, five. The line 𝐿 is a tangent to the circle of equation π‘₯ squared plus 𝑦 squared is equal to 34 at the point 𝑃. The line 𝐿 crosses the π‘₯-axis at the point 𝑀. Calculate the area of triangle 𝑂𝑃𝑀.

Let’s start by drawing a sketch to help us answer this question. We can recall the general equation of a circle with a center at the origin and a radius π‘Ÿ to be π‘₯ squared plus 𝑦 squared equals π‘Ÿ squared. That means then that we have a circle with a center at the origin.

To find the radius of our circle, we can form an equation. Since π‘Ÿ squared in the general form for the equation of a circle is 34 in our equation, π‘Ÿ squared must be equal to 34. We can square root both sides of this equation to give us a radius of root 34.

Next, we need to add the coordinate 𝑃. Remember the diagram doesn’t need to be to scale. But everything should be roughly in the right place. We can plot the coordinate negative three, five in the second quadrant. Next, we add the tangent at the point 𝑃 and the point 𝑀, where our tangent crosses the π‘₯-axis.

Now, let’s look back to the question. We’re being asked to find the area of triangle 𝑂𝑃𝑀. That’s this one highlighted on our diagram. Remember the formula for the area of a triangle is a half multiplied by the length of the base multiplied by its height. So let’s see if we know any of these dimensions. First, we’ll need to decide which is the height and which is the base of the triangle.

If we use the line 𝑂𝑀 as the base, we can really quickly identify the height of the triangle. The height is given by this dotted line. Since we know the dotted line is vertical and it starts at the π‘₯-axis, we can use the 𝑦-coordinate of point 𝑃 to show us that the height of the triangle is five units.

But what about the length of its base? Well, since point 𝑂 is at the origin, if we can find the π‘₯-coordinate of point 𝑀, that will tell us the width of this triangle. In order to do this, we’ll first need to find the equation of the line 𝐿.

The equation for a straight line is given by 𝑦 is equal to π‘šπ‘₯ plus 𝑐, where π‘š is the gradient and 𝑐 is the 𝑦-intercept. In order to find the gradient of the tangent, we must first find the gradient of the radius. The formula for the gradient of a straight line is change in 𝑦 over change in π‘₯, which is sometimes written as 𝑦 two minus 𝑦 one over π‘₯ two minus π‘₯ one. Don’t be scared of these little numbers. It’s just a way of reminding us to always perform subtractions in the same order.

Let’s see what that looks like. We first find two coordinates that lie on the radius of the circle: zero, zero and negative three, five. If we take zero, zero to be π‘₯ one, 𝑦 one and negative three, five to be π‘₯ two, 𝑦 two, then we can substitute these values into the formula for the gradient. That gives us five minus zero divided by negative three minus zero. That simplifies to negative five-thirds.

It would’ve have also been fine to make negative three, five be π‘₯ one, 𝑦 one and zero, zero be π‘₯ two, 𝑦 two. Substituting these values into the formula for the gradient gives us zero minus five divided by zero minus negative three, which is also negative five-thirds.

Now that we know the gradient of the radius, we can calculate the gradient of the tangent. The radius and the tangent meet at a right angle. That means they’re perpendicular. To find the gradient of a line that is perpendicular, we find the negative reciprocal of the gradient of the original line. In this case, the reciprocal of five-thirds is three-fifths. It was negative to begin with, which means the gradient of the tangent is positive three-fifths.

Another way to remember this is that the gradients of two lines are perpendicular must multiply to make negative one. Let’s calculate negative five-thirds multiplied by three-fifths. Then, we can cross cancel by dividing through by a common factor of three and five. Negative five-thirds multiplied by three-fifths is negative one. This can be a really nice way to check the answer is correct.

Now that we know the gradient of our tangent, let’s substitute it into the formula for a straight line. We do know one more piece of information. And that’s that the point negative three, five lies on the tangent. That means we can substitute π‘₯ equals negative three and 𝑦 equals five into this equation, giving us five is equal to three-fifths multiplied by negative three plus 𝑐.

Let’s multiply three-fifths by negative three by writing negative three as negative three over one. Three multiplied by negative three is negative nine and five multiplied by one is five. Our equation becomes five is equal to negative nine-fifths plus 𝑐.

Next, we’ll solve this equation by adding nine-fifths to both sides. Once again, let’s write five as five over one. Then, we can find a common denominator to add these fractions. The lowest common denominator of these two fractions is five. So we’ll multiply the numerator and the denominator of five over one by five. That gives us twenty-five fifths. Twenty-five fifths plus nine-fifths is equal to thirty-four fifths.

Now, we have the equation for our line 𝐿. It’s 𝑦 is equal to three-fifths π‘₯ plus thirty-four fifths. But where does this line cross the π‘₯-axis? Well, an alternative way of labelling the π‘₯-axis is to call it the line 𝑦 equals zero since all coordinates on this line have a 𝑦-coordinate of zero. That means we can replace the 𝑦 with zero in our equation for the tangent and then solve to find π‘₯.

Let’s subtract thirty-four fifths from both sides of this equation. That gives us negative thirty-four fifths is equal to three-fifths π‘₯. Since both sides of this equation have the same denominator, this means their numerators must be equal. Negative 34 is equal to three π‘₯.

Finally, let’s divide both sides of this equation by three. That tells us that the π‘₯-coordinate of the point 𝑀 is negative thirty-four thirds. The length of the base of the triangle then is thirty-four thirds units. Remember this can’t be negative because it’s a measure of length. Now, we have all the information we need to find the area of our triangle.

Substituting the values we know into the formula for the area of a triangle give us a half multiplied by five multiplied by thirty-four thirds. We can write five as five over one and cross cancel by dividing through by a common factor of two. One multiplied by five multiplied by 17 is 85 and one multiplied by one multiplied by three is three.

Let’s change eighty-five thirds into a mixed number by dividing 85 by three. Eight divided by three is two remainder two and 25 divided by three is eight remainder one. This means that eighty-five thirds is the same as twenty-eight and a third. And the area of triangle 𝑂𝑃𝑀 is twenty-eight and a third units squared.

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