### Video Transcript

π is the point negative three,
five. The line πΏ is a tangent to the
circle of equation π₯ squared plus π¦ squared is equal to 34 at the point π. The line πΏ crosses the π₯-axis at
the point π. Calculate the area of triangle
πππ.

Letβs 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 to be π₯ squared plus π¦
squared equals π 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 squared in the general
form for the equation of a circle is 34 in our equation, π 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
π. Remember the diagram doesnβt 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 π and the point π, where our tangent crosses the π₯-axis.

Now, letβs look back to the
question. Weβre being asked to find the area
of triangle πππ. Thatβs 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 letβs see if we know any of
these dimensions. First, weβll need to decide which
is the height and which is the base of the triangle.

If we use the line ππ 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 π¦-coordinate of point π to
show us that the height of the triangle is five units.

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

The equation for a straight line is
given by π¦ is equal to ππ₯ plus π, where π is the gradient and π is the
π¦-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 π¦ over change in π₯, which is sometimes written as π¦
two minus π¦ one over π₯ two minus π₯ one. Donβt be scared of these little
numbers. Itβs just a way of reminding us to
always perform subtractions in the same order.

Letβs 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,
π¦ one and negative three, five to be π₯ two, π¦ 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 wouldβve have also been fine to
make negative three, five be π₯ one, π¦ one and zero, zero be π₯ two, π¦ 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 theyβre
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. Letβs 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, letβs substitute it into the formula for a straight line. We do know one more piece of
information. And thatβs that the point negative
three, five lies on the tangent. That means we can substitute π₯
equals negative three and π¦ equals five into this equation, giving us five is equal
to three-fifths multiplied by negative three plus π.

Letβs 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 π.

Next, weβll solve this equation by
adding nine-fifths to both sides. Once again, letβs 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 weβll 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 πΏ. Itβs π¦ 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 π¦ equals zero since all coordinates on
this line have a π¦-coordinate of zero. That means we can replace the π¦
with zero in our equation for the tangent and then solve to find π₯.

Letβs 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, letβs divide both sides of
this equation by three. That tells us that the
π₯-coordinate of the point π is negative thirty-four thirds. The length of the base of the
triangle then is thirty-four thirds units. Remember this canβt be negative
because itβs 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.

Letβs 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 πππ is
twenty-eight and a third units squared.