# Video: CBSE Class X • Pack 2 • 2017 • Question 28

CBSE Class X • Pack 2 • 2017 • Question 28

04:57

### Video Transcript

If the points 𝐴, located at 𝐾 plus one, two 𝐾; 𝐵, located at three 𝐾, two 𝐾 plus three; and 𝐶, located at five 𝐾 minus one, five 𝐾 are collinear, find the value of 𝐾.

So if 𝐴, 𝐵, and 𝐶, all three points, are collinear, then they lie on the same line. The problem is we don’t know the equation of this line. However, if they are all on the same line, then they all should have the same slope. And the formula to find the slope of a line is 𝑦 two minus 𝑦 one divided by 𝑥 two minus 𝑥 one. So if 𝐴, 𝐵, and 𝐶 all lie on the same line, this means the slope of 𝐴𝐵 should be the same as the slope of 𝐵𝐶 which should be equal to the slope of 𝐴𝐶. So let’s find each of these slopes. And then set them equal to each other. And then solve for 𝐾.

So here are our points. And let’s find the slope of 𝐴𝐵. 𝐴 can be 𝑥 one, 𝑦 one. And 𝐵 can be 𝑥 two, 𝑦 two. So we have two 𝐾 plus three minus two 𝐾 divided by three 𝐾 minus 𝐾 plus one. So we need to be careful with the minus signs. On the numerator, we simply have minus two 𝐾. But on the denominator, we have minus 𝐾 minus one. So now let’s simplify. On the numerator, the two 𝐾s cancel. And we have three divided by two 𝐾 minus one. Three 𝐾 minus 𝐾 on the denominator is equal to two 𝐾. So here we have the slope of 𝐴𝐵.

Now, let’s find the slope of 𝐵𝐶. 𝐵 can be 𝑥 one, 𝑦 one and 𝐶 can be 𝑥 two, 𝑦 two. So we have five 𝐾 minus two 𝐾 plus three. And then on the denominator, five 𝐾 minus one minus three 𝐾. On the numerator, we distribute the negative sign. And we have minus two 𝐾 minus three. And on the denominator, we distribute the negative. And it’s just minus three 𝐾. So after combining like terms, we have that the slope of 𝐵𝐶 is three 𝐾 minus three over two 𝐾 minus one.

Lastly, we find the slope of 𝐴𝐶. 𝐴 can be 𝑥 one, 𝑦 one and 𝐶 can be 𝑥 two, 𝑦 two. On the numerator, we have five 𝐾 minus two 𝐾. And on the denominator, we have five 𝐾 minus one minus 𝐾 plus one. So on the numerator, we have five 𝐾 minus two 𝐾. And on the denominator, we need to distribute the negative. And now, we simplify. And we find that the slope of 𝐴𝐶 is equal to three 𝐾 over four 𝐾 minus two. So now, to solve for 𝐾, we need to set the slopes equal to each other. So we can essentially just pick two of them and set them equal to each other and be done. However, to be extra safe, we could set a few of them equal to each other. First, let’s set these two equal to each other.

So to solve, we can essentially cross multiply, or multiplying both sides on the numerator by two 𝐾 minus one. However, this just makes the two 𝐾 minus ones cancel. So we have that three is equal to three 𝐾 minus three. So we add three to both sides of the equation, divide both sides by three, and find that 𝐾 is equal to two. So we found that 𝐾 is equal to two. Now, let’s set another two of the slopes equal to each other. We should get the same answer of 𝐾 equals two.

So we’re setting the slope of 𝐵𝐶 equal to the slope of 𝐴𝐶 right now. So we can cross multiply. Or, we can multiply both sides by two 𝐾 minus one to get rid of it on the denominator. And then multiply by both sides of four 𝐾 minus two to get rid of it on the denominator. Let’s simply cross multiply. So we have three 𝐾 times two 𝐾 minus one. And we have three 𝐾 minus three times four 𝐾 minus two.

To make this a little easier, four 𝐾 minus two can be changed by taking out a GCF of two, a greatest common factor. And this would make the two 𝐾 minus ones cancel. So now, we would need to distribute the two. And we would have three 𝐾 equals six 𝐾 minus six. So if we would subtract six 𝐾 from both sides of the equation, we would end up with negative three 𝐾 equals negative six. And if we would solve for 𝐾, we will divide both sides of the equation by negative three and find that 𝐾 is equal to two, just as we found before.

And now lastly, if we wanted to triple check, we could set the slope of 𝐴𝐵 equal to the slope of 𝐴𝐶. This one is set up similar to the one that we just did. Because four 𝐾 minus two could be rewritten as two times two 𝐾 minus one. So when we would cross multiply, we would have three 𝐾 times two 𝐾 minus one and three times two times two 𝐾 minus one. The two 𝐾 minus ones would cancel. And we would have that three 𝐾 is equal to three times two. And three times two is six. So after dividing both sides of the equation by three, we find that 𝐾 is equal to two. Therefore, we can be very sure that if 𝐴, 𝐵, and 𝐶 are collinear, the value of 𝐾 would be two.