What are the coordinates of P if it divides ad in the ratio 2 1?

Suppose you have a line segment PQ ¯ on the coordinate plane, and you need to find the point on the segment 1 3  of the way from P to Q .

Let’s first take the easy case where P is at the origin and line segment is a horizontal one.

The length of the line is 6 units and the point on the segment 1 3  of the way from P to Q would be 2 units away from P , 4 units away from Q and would be at ( 2,0 ) .

Consider the case where the segment is not a horizontal or vertical line.

The components of the directed segment PQ ¯ are 〈 6,3 〉  and we need to find the point, say X on the segment 1 3  of the way from P to Q .

Then, the components of the segment PX ¯ are 〈 ( 1 3 )( 6 ),( 1 3 )( 3 ) 〉=〈 2,1 〉 .

Since the initial point of the segment is at origin, the coordinates of the point X are given by ( 0+2,0+1 )=( 2,1 ) .

Now let’s do a trickier problem, where neither P nor Q is at the origin.

Use the end points of the segment PQ ¯  to write the components of the directed segment.

〈 ( x 2 − x 1 ),( y 2 − y 1 ) 〉=〈 ( 7−1 ),( 2−6 ) 〉                                             =〈 6,−4 〉

Now in a similar way, the components of the segment PX ¯  where X is a point on the segment 1 3  of the way from P to Q are 〈 ( 1 3 )( 6 ),( 1 3 )( −4 ) 〉=〈 2,−1.25 〉 .

To find the coordinates of the point X add the components of the segment PX ¯  to the coordinates of the initial point P .

So, the coordinates of the point X are ( 1+2,6−1.25 )=( 3,4.75 ) .

Note that the resulting segments, PX ¯ and XQ ¯ , have lengths in a ratio of 1:2 .

In general: what if you need to find a point on a line segment that divides it into two segments with lengths in a ratio a:b ?

Consider the directed line segment XY ¯  with coordinates of the endpoints as X( x 1 , y 1 )  and Y( x 2 , y 2 ) .

Suppose the point Z divided the segment in the ratio a:b , then the point is a a+b of the way from X to Y .

So, generalizing the method we have, the components of the segment XZ ¯ are 〈 ( a a+b ( x 2 − x 1 ) ),( a a+b ( y 2 − y 1 ) ) 〉 .

Then, the X -coordinate of the point Z is

x 1 + a a+b ( x 2 − x 1 )= x 1 ( a+b )+a( x 2 − x 1 ) a+b                                         = b x 1 +a x 2 a+b .

Similarly, the Y -coordinate is

y 1 + a a+b ( y 2 − y 1 )= y 1 ( a+b )+a( y 2 − y 1 ) a+b                                         = b y 1 +a y 2 a+b .

Therefore, the coordinates of the point Z are ( b x 1 +a x 2 a+b , b y 1 +a y 2 a+b ) .

Example 1:

Find the coordinates of the point that divides the directed line segment MN ¯ with the coordinates of endpoints at M( −4,0 )  and M( 0,4 ) in the ratio 3:1 ?

Let L be the point that divides MN ¯  in the ratio 3:1 .

Here, ( x 1 , y 1 )=( −4,0 ),( x 2 , y 2 )=( 0,4 ) and a:b=3:1 .

Substitute in the formula. The coordinates of L are

( 1( −4 )+3( 0 ) 3+1 , 1( 0 )+3( 4 ) 3+1 ) .

Simplify.

( −4+0 4 , 0+12 4 )=( −1,3 )

Therefore, the point L( −1,3 )  divides MN ¯  in the ratio 3:1 .

Example 2:

What are the coordinates of the point that divides the directed line segment AB ¯  in the ratio 2:3 ?

Let C be the point that divides AB ¯  in the ratio 2:3 .

Here, ( x 1 , y 1 )=( −4,4 ),( x 2 , y 2 )=( 6,−5 ) and a:b=2:3 .

Substitute in the formula. The coordinates of C are

( 3( −4 )+2( 6 ) 5 , 3( 4 )+2( −5 ) 5 ) .

Simplify.

( −12+12 5 , 12−10 5 )=( 0, 2 5 )                                             =( 0,0.4 )

Therefore, the point C( 0,0.4 )  divides AB ¯  in the ratio 2:3 .

You can note that the Midpoint Formula is a special case of this formula when a=b=1 .

Solution:

Let the coordinates of the point be P(x, y) which divides the line segment joining the points (-1, 7) and (4, - 3) in the ratio 2 : 3

Let two points be A (x₁, y₁) and B(x₂, y₂). P (x, y) divides internally the line joining A and B in the ratio m₁: m₂. Then, coordinates of P(x, y) is given by the section formula

P (x, y) = [(mx₂ + nx₁ / m + n), (my₂ + ny₁ / m + n)]

Let  x₁ = - 1, y₁ = 7, x₂ = 4 and y₂ = - 3, m = 2, n = 3

By Section formula, P (x, y) = [(mx₂ + nx₁ / m + n) , (my₂ + ny₁ / m + n)] --- (1)

By substituting the values in the equation (1)

x = [2 × 4 + 3 × (- 1)] / (2 + 3) and y = [2 × (- 3) + 3 × 7] / (2 + 3)

x = (8 - 3) / 5 and  y = (- 6 + 21) / 5

x = 5/5 = 1 and y = 15/5 = 3

Therefore, the coordinates of point P are (1, 3).

☛ Check: NCERT Solutions Class 10 Maths Chapter 7

Video Solution:

Find the coordinates of the point which divides the join of (-1, 7) and (4, - 3) in the ratio 2 : 3

NCERT Class 10 Maths Solutions Chapter 7 Exercise 7.2 Question 1

Summary:

The coordinates of the point which divides the join of (- 1, 7) and (4, - 3) in the ratio 2 : 3 is (1, 3).

☛ Related Questions:

• Find the coordinates of the points of trisection of the line segment joining (4, - 1) and (- 2, - 3).
• To conduct Sports Day activities, in your rectangular-shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1m each. 100 flower pots have been placed at a distance of 1m from each other along with AD, as shown in Fig. 7.12. Niharika runs 1/4 th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5 th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segments joining the two flags, where should she post her flag?
• Find the ratio in which the line segment joining the points (- 3, 10) and (6, - 8) is divided by (- 1, 6).
• Find the ratio in which the line segment joining A (1, - 5) and B (- 4, 5) is divided by the x-axis. Also, find the coordinates of the point of division.

Given two points \(A\), \(B\) in the plane, it is clearly possible to create a number line on \(AB\) so as to label each point on \(AB\) with a (real) number.

There are (infinitely) many ways in which this can be done, but it turns out not to be particularly useful for geometrical purposes.

On the other hand, in the following diagrams, we can say that the point \(P\) divides the interval \(AB\) in the ratio \(1:1\) and the point \(Q\) divides the interval \(AB\) in the ratio \(3:2\).

Midpoint of an interval

The midpoint of an interval \(AB\) is the point that divides \(AB\) in the ratio \(1:1\).

Assume that the point \(A\) has coordinates \((x_1,y_1)\) and the point \(B\) has coordinates \((x_2,y_2)\). It is easy to see, using either congruence or similarity, that the midpoint \(P\) of \(AB\) is

This is shown in the module Introduction to coordinate geometry (Years 9–10).

Internal division of an interval

We now generalise the idea of a midpoint to that of a point that divides the interval \(AB\) in the ratio \(k:1\).

Definition

Suppose \(k>0\) is a real number and let \(P\) be a point on a line interval \(AB\). Then \(P\) divides \(AB\) in the ratio \(k:1\) if

Note. As \(k\to 0\), \(P\to A\) and as \(k\to \infty\), \(P\to B\).

Theorem

Let \(A(x_1,y_1)\) and \(B(x_2,y_2)\) be two points in the plane and let \(P(x,y)\) be the point that divides the interval \(AB\) in the ratio \(k:1\), where \(k>0\). Then

Proof

If \(x_1 = x_2\), then it is clear that \(x\) is given by the formula above. So we can assume that \(x_1 \ne x_2\). Consider the points \(C(x, y_1)\) and \(D(x_2, y_1)\), as shown in the following diagram.

The triangles \(ACP\) and \(ADB\) are similar (AAA). So \begin{align*} \dfrac{AP}{AB} &= \dfrac{AC}{AD}\\ \dfrac{k}{k+1} &= \dfrac{x-x_1}{x_2-x_1}\\ kx_2-kx_1 &= kx-kx_1+x-x_1\\ (k+1)x &= kx_2+x_1\\ x &= \dfrac{x_1+kx_2}{1+k}. \end{align*}
Similarly, \[ y = \dfrac{y_1+ky_2}{1+k}. \]

\(\Box\)

Exercise 1

If \(P(x,y)\) lies on the interval \(A(x_1,y_1)\), \(B(x_2,y_2)\) such that \(AP:PB = a:b\), with \(a\) and \(b\) positive, show that

Screencast of exercise 1

The formulas of Exercise 1 are worth learning.

Let \(A\) be \((-3,5)\) and \(B\) be \((5,-10)\). Find

1. the distance \(AB\)
2. the midpoint \(P\) of \(AB\)
3. the point \(Q\) which divides \(AB\) in the ratio \(2:5\).

Solution

1. \(AB^2 = \bigl(5-(-3)\bigr)^2+\bigl(-10-5\bigr)^2 = 8^2+15^2 = 17^2\), so \(AB=17\).
2. \(P\) has coordinates \(\bigl(1,-\dfrac{5}{2}\bigr)\).
3. \(Q\) has coordinates \[ \Bigl(\dfrac{5\times-3 + 2\times5}{5+2}, \dfrac{5\times5 + 2\times-10}{5+2}\Bigr) = \Bigl(-\dfrac{5}{7}, \dfrac{5}{7}\Bigr). \]

Exercise 2

Give an alternative proof of the previous theorem on internal division of an interval. Use the fact that \(AC = AP\cos\theta\) in the diagram from the proof, where \(\theta = \angle PAC\).

Exercise 3

Suppose points \(A\), \(B\) and \(C\) are not collinear and have coordinates \((x_1,y_1)\), \((x_2,y_2)\) and \((x_3,y_3)\). Let \(D\) be the midpoint of \(BC\) and suppose \(G\) divides the median \(AD\) in the ratio \(2:1\). Find the coordinates of \(G\) and deduce that the medians of \(\triangle ABC\) are concurrent.

External division of an interval

Dividing an interval \(AB\) internally in a given ratio produces a point between \(A\) and \(B\). External division produces a point outside the interval \(AB\).

Suppose \(D\), \(A\), \(B\) and \(C\) are collinear and \(DA=AB=BC\), as in the above diagram. Then

We say that \(C\) divides \(AB\) externally in the ratio \(2:1\) and that \(D\) divides \(AB\) externally in the ratio \(1:2\). Clearly this is different from the internal division of an interval discussed in the previous subsection.

In general, suppose that \(P(x,y)\) is on the line \(AB\) but is external to the interval \(AB\) and that \(\dfrac{AP}{PB} = \dfrac{k}{1}\), for some \(k > 0\).

Then

This suggests that we make the following definition.

Definition

Suppose \(k < 0\) is a real number with \(k \ne -1\) and let \(P\) be a point on a line \(AB\). Then \(P\) divides the interval \(AB\) in the ratio \(k:1\) if \(P\) is external to the interval and

With this convention we have

for external division, exactly as for the internal division of an interval! These formulas are algebraically the same as for internal division, but here \(k\) is negative.

Find the coordinates of the point \(P\) which divides the interval \(A(-3,-7)\), \(B(-1,-4)\) externally in the ratio \(4:3\).

Solution

Here \(k = -\dfrac{4}{3}\) and

The point \(P\) has coordinates \((5, 5)\).

Now, for each point \(P\) on the line \(AB\), we have an associated number \(k\):

• for \(P\) in the interval \(AB\), we take \(k=\dfrac{AP}{PB}\)
• for \(P\) on the line \(AB\) external to the interval \(AB\), we take \(k=-\dfrac{AP}{PB}\).

The following diagram shows the values of \(k\) for the marked points.

Clearly, as \(P\to\infty\) in either direction, \(\dfrac{AP}{PB}\to -1\). (We are now close to a concept called the real projective line.)

Exercise 4

Suppose \(M\) is the midpoint of \(AB\), where \(A\) is \((8,10)\) and \(B\) is \((18,20)\). Further suppose that \(P\) divides \(AB\) internally and \(Q\) divides \(AB\) externally in the ratio \(2:3\). Show that \(MP \cdot MQ = MB^2\).

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