# Video: CBSE Class X • Pack 5 • 2014 • Question 20

CBSE Class X • Pack 5 • 2014 • Question 20

03:32

### Video Transcript

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 equations.

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 𝐴𝐵.

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 the three.

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.

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