If the points 𝐴 negative two, one,
𝐵 𝑎, 𝑏, and 𝐶 four, negative one are collinear and a minus 𝑏 equals one, find
the values of 𝑎 and 𝑏.
So we’re told that the points 𝐴,
𝐵, and 𝐶 are collinear, that means they lie on the same line, so something like
this. And we’re asked to find the values
of 𝑎 and 𝑏, which are the 𝑥- and 𝑦-coordinate of point 𝐵. We need to create some
Well, since points 𝐴, 𝐵, and 𝐶
are collinear lying on the same line, then the segments joining them will have the
same slope. Slope 𝑚 is a change in 𝑦 divided
by the change in 𝑥, which is the rise divided by the run. So the slope of segment 𝐴𝐵 will
be equal to the slope of 𝐵𝐶, which will be equal to the slope of 𝐴𝐶. This is because 𝐴, 𝐵, and 𝐶
those points are collinear.
And since we know the location of
𝐴 and 𝐶, let’s find the slope of segment 𝐴𝐶. 𝐴 is negative two, one and 𝐶 is
four, negative one. So we can let 𝐴 be 𝑥 one, 𝑦 one
and 𝐶 be 𝑥 two, 𝑦 two. So plugging in our values, on our
numerator negative one minus one is negative two and four minus negative two is
really four plus two which is six. And this reduces to negative
one-third because two and six are both divisible by two.
So if we know the slope of 𝐴𝐶 and
it’s negative one-third, then this slope should also be negative one-third. So if we’re looking for the
coordinates of 𝐵, let’s find the slope equation with the point 𝐵 in it and set it
equal to one-third. So let’s find the slope of
Let’s plug in our values. On the numerator, we have 𝑏 minus
one, which doesn’t simplify. But on the denominator, 𝑎 minus
negative two is really 𝑎 plus two. So the slope of 𝐴𝐵 is equal to 𝑏
minus one divided by 𝑎 plus two. But it’s also equal to negative
one-third. So let’s set these equal to each
other and create an equation in terms of 𝑎 and 𝑏.
To begin, let’s get rid of the
three on the denominator. Let’s multiply both sides of the
equation by three. On the right, the threes
cancel. On the left, we need to distribute
Now, let’s get rid of 𝑎 plus two
on the bottom, which we will do so by multiplying both sides of the equation by 𝑎
plus two. On the left, they cancel and on the
right, we distribute.
So here we have an equation with 𝑎
and 𝑏. We can’t solve for 𝑎 or 𝑏 because
we have two variables. However, in the question, we were
given that 𝑎 minus 𝑏 is equal to one. So we can take this equation and
solve for 𝑎. So we can add 𝑏 to both sides. So we also know that 𝑎 is equal to
𝑏 plus one. So we can substitute 𝑏 plus one in
for 𝑎. So here, we’ve replaced 𝑎 with 𝑏
plus one. We’ve distributed the negative.
And now, let’s combine negative one
and negative two to get negative three. So now, let’s add 𝑏 to both sides
of the equation. Now, we need to add three to both
sides of the equation. And we have that four 𝑏 is equal
to zero. So now we divide both sides by
four, resulting in 𝑏 equals zero. So now that we know 𝑏, we can plug
it back into the equation and solve for 𝑎. So 𝑎 will be equal to zero plus
one which is one.
Therefore, 𝑎 is equal to one and
𝑏 is equal to zero, which means 𝐵 is located at one, zero.