If 𝐴: 𝑥, two; 𝐵: negative three,
negative four; and 𝐶: seven, negative five are collinear, find the value of 𝑥.
So if points are collinear, that
means they lie on the same line. So we could have something like
this. And because they’re collinear,
lying on the same line, the segments joining these points will have the same
Slope 𝑚 is the change in 𝑦, which
is the rise, divided by the change in 𝑥, the run. So if we found the slope of segment
𝐵𝐶, we could use it to find the equation of the line between all three points. And then once we found the equation
of the line, we can plug in one of the values of 𝐴, which is two — we’re given the
𝑦-value as two. We can plug that into the equation
of the line and then find 𝑥.
So let’s begin by finding the slope
of segment 𝐵𝐶. So our formula to find a slope is
the change in 𝑦 divided by the change in 𝑥. The change in 𝑦 is where we will
take 𝑦 two and then subtract 𝑦 one and then divide by 𝑥 two minus 𝑥 one.
So to find the slope of segment
𝐵𝐶, we will let the point 𝐵 be 𝑥 one, 𝑦 one and the point 𝐶 𝑥 two, 𝑦
two. So on the numerator, we have
negative five minus negative four, giving us negative one. And on the denominator, seven minus
negative three, that gives us positive 10. So the slope of segment 𝐵𝐶 is
And we know that the slope of 𝐵𝐶
will be the exact same as the slope of segment 𝐴𝐵, which would be the exact same
slope as segment 𝐴𝐶 because they are all colinear; they all lie on the same line,
So now we need to solve for 𝑏. 𝑏- 𝑏 is the 𝑦-intercept where
crosses the 𝑦-axis, where our line crosses the 𝑦-axis. And in order to find 𝑏, we need to
plug an 𝑥 and a 𝑦 in. We can use point 𝐵 or point 𝐶 to
plug in. Let’s use point 𝐵.
So 𝐵 is negative three, negative
four. So negative three is 𝑥, and
negative four is 𝑦. Let’s substitute those into our
equation of the line and solve for 𝑏. So here we’ve done so.
So let’s first begin by multiplying
negative one-tenth times negative three. That gives us positive
three-tenths. So now, so we have positive three
tenths, and let’s get rid of the denominator. Let’s multiply everything by
10. This way, on the three tenths, the
10s cancel and there’s no more denominator.
So to solve for 𝑏, let’s subtract
three from both sides of the equation. The threes cancel, and now we need
to divide both sides of the equation by 10. This means that 𝑏 is equal to
negative forty-three tenths.
So now we have the equation of this
line that all of the points 𝐴, 𝐵, and 𝐶 lie on. 𝑦 equals negative one tenth 𝑥
minus forty-three tenths. And since we know that point 𝐴
lies on this line, we can plug in two for 𝑦 and solve for the missing value 𝑥.
So let’s replace 𝑦 with two. And this will allow us to solve for
𝑥, the missing 𝑥 value of point 𝐴. So let’s begin by multiplying
everything by 10, just like we did before to get rid of the denominators, resulting
in 20 equals negative 𝑥 minus 43.
So now we need to add 43 to both
sides of the equation, and we have 63 equals negative 𝑥. But we want 𝑥. So we need to divide both sides of
the equation by negative one, resulting in 𝑥 equals negative 63, which would be our
However, let’s think about
this. That’s kind of a big number. Let’s go ahead and plot 𝐵 and 𝐶
and then determine where 𝐴 should fall. Point 𝐵 is here at negative three,
negative four. Point 𝐶 is here at seven, negative
five. And we know that they’re collinear,
so they’re all on the same line. And we know that 𝐴 is at 𝑥 comma
two. So 𝑦 is two, which means 𝐴 also
lies somewhere on this line.
And while this sketch is
inaccurate, it does show that 𝐴 needs to be very far to the left on the
𝑥-axis. So finding that the 𝑥 value for 𝐴
is negative 63 actually seems pretty good. So once again, the value of 𝑥 is
equal to negative 63.