# Video: CBSE Class X • Pack 3 • 2016 • Question 6

CBSE Class X • Pack 3 • 2016 • Question 6

07:06

### Video Transcript

Let 𝑃 and 𝑄 be the points of trisection of the line segment joining the points 𝐴: two, negative two and 𝐵: negative seven, four, such that 𝑃 is nearer to 𝐴. Find the coordinates of 𝑃 and 𝑄.

There are several ways of solving this problem. We could do it graphically or we could do it using the section formula. Our first method will consider the line 𝐴𝐵 from two, negative two to 𝐵: negative seven, four that has been split into three equal sections, the trisections, by the points 𝑃 and 𝑄.

If we consider the 𝑥-coordinates of 𝐴 and 𝐵, they are two and negative seven. To get from two to negative seven, we need to subtract nine. Negative nine divided by three is equal to negative three. This means that the 𝑥-coordinate at each point will be three less than the previous one. Two minus three is equal to negative one. Therefore, the 𝑥-coordinate of 𝑃 is negative one. Negative one minus three is equal to negative four. Therefore, the 𝑥-coordinate of 𝑄 is negative four. Subtracting three from negative four gives us negative seven, which is the 𝑥-coordinate of 𝐵.

We can then use the same method for the 𝑦-coordinate. To get from negative two to four, we need to add six. Once again, there are three jumps. Six divided by three is equal to two. Therefore, we need to add two to each of the 𝑦-coordinates. Negative two plus two is equal to zero. Therefore, the 𝑦-coordinate of 𝑃 is zero. Zero plus two is equal to two. Therefore, the 𝑦-coordinate of 𝑄 is two. Finally, by adding two to two gives us an answer of four. The 𝑦-coordinate of 𝐵 is four. We can therefore say that the points of trisection 𝑃 and 𝑄 have coordinates negative one, zero and negative four, two.

An alternative method would be to plot the points accurately on a coordinate axes. Point 𝐴 had coordinates two, negative two and point 𝐵 had coordinates negative seven, four. To get from point 𝐴 to point 𝐵, we move left nine units and up six units.

As we’re looking for the points of trisection, we can divide these values by three. Nine divided by three is equal to three, and six divided by three is equal to two. Starting at point 𝐴 and moving three units left and two units up takes us to point 𝑃. This has coordinates negative one, zero. Moving another three units left and two units up takes us to point 𝑄. This has coordinates negative four, two. Finally, moving three units left and two units up from point 𝑄 takes us to point 𝐵. This method also gives us the points of trisection negative one, zero for 𝑃 and negative four, two for 𝑄.

The third method we will look at to solve this problem is using the section formula. If point 𝑃 lies on the line segment 𝐴𝐵 between points 𝐴 and 𝐵 and satisfies the ratio 𝐴𝑃 to 𝑃𝐵 is equal to 𝑚 to 𝑛, where the coordinate of 𝐴 is 𝑥 one, 𝑦 one and the coordinate of 𝐵 is 𝑥 two, 𝑦 two, then point 𝑃 has coordinates 𝑚𝑥 two plus 𝑛𝑥 one divided by 𝑚 plus 𝑛 and 𝑚𝑦 two plus 𝑛𝑦 one divided by 𝑚 plus 𝑛.

Let’s consider our question where 𝐴 has coordinates two, negative two and 𝐵 has coordinates negative seven, four. As point 𝑃 was a point of trisection, the ratio of 𝐴𝑃 to 𝑃𝐵 is one to two. Substituting these values into the formula gives us an 𝑥-coordinate of 𝑃 of one multiplied by negative seven plus two multiplied by two divided by one plus two and a 𝑦-coordinate of one multiplied by four plus two multiplied by negative two divided by one plus two.

Simplifying this gives us an 𝑥-coordinate of negative seven plus four divided by three and a 𝑦-coordinate of four minus four divided by three. Negative seven plus four is equal to negative three, and negative three divided by three is equal to negative one. Therefore, the 𝑥-coordinate of 𝑃 is negative one. Four minus four is equal to zero, and zero divided by three is zero. Therefore, the 𝑦-coordinate of 𝑃 is equal to zero. This confirms that the coordinate of 𝑃 is negative one, zero.

We can use the same method to work out the coordinates of point 𝑄. This time, the ratio of 𝐴𝑄 to 𝑄𝐵 is two to one. This means that the 𝑥-coordinate of 𝑄 can be calculated by multiplying two by negative seven plus one multiplied by two divided by three. The 𝑦-coordinate can be calculated by multiplying two by four plus one multiplied by negative two divided by three.

The 𝑥-coordinate can be simplified to negative 14 plus two divided by three and a 𝑦-coordinate to eight minus two divided by three. Negative 14 plus two is negative 12. Dividing this by three gives us an 𝑥-coordinate of negative four. Eight minus two is equal to six, and dividing this by three gives us a 𝑦-coordinate of two. This means that the coordinates of point 𝑄 are negative four, two, giving us the same answers as in our first two methods.