Find the ratio in which the line segment joining the points and is divided by .

Understand the Question Let the point divide the line segment joining and internally in the ratio . By the section formula, the coordinates of a point dividing the line segment between and in the ratio are given by: We can solve for using the -coordinate and verify it using the -coordinate. Step 1 · Find the Ratio using x-coordinate Let the point divide the segment joining and in the ratio .Using the section formula for the -coordinate: Substitute , , and : Thus, the required ratio is . Step 2 · Verify with y-coordinate Using the section formula for the -coordinate: Substitute , , and : Both coordinates yield . Answer > Common Mistakes Swapping Point Order: Taking first and second gives the ratio instead of . The ratio must correspond to the directed line segment from to . Sign Error in Distribution: When expanding , mistakenly writing instead of . Skipping Verification: It is important to check the ratio in the -coordinate equation to confirm that the three points are indeed collinear.

Coordinate Geometry | Exercise 7.2

Question 4

Find the ratio in which the line segment joining the points $(-3, 10)$ and $(6, -8)$ is divided by $(-1, 6)$ .

Solution

Understand the Question

Let the point $P(-1, 6)$ divide the line segment joining $A(-3, 10)$ and $B(6, -8)$ internally in the ratio $k : 1$ .
By the section formula, the coordinates of a point $P(x, y)$ dividing the line segment between $A(x_1, y_1)$ and $B(x_2, y_2)$ in the ratio $m : n$ are given by: $x = \dfrac{m x_2 + n x_1}{m + n}, \quad y = \dfrac{m y_2 + n y_1}{m + n}$
We can solve for $k$ using the $x$ -coordinate and verify it using the $y$ -coordinate.

Step 1 · Find the Ratio using x-coordinate

Let the point $P(-1, 6)$ divide the segment joining $A(-3, 10)$ and $B(6, -8)$ in the ratio $k : 1$ .Using the section formula for the $x$ -coordinate: $x = \dfrac{k x_2 + 1 \cdot x_1}{k + 1}$

Substitute $(x_1, y_1) = (-3, 10)$ , $(x_2, y_2) = (6, -8)$ , and $x = -1$ :

\begin{aligned} -1 &= \dfrac{k(6) + 1(-3)}{k + 1} \\[0.6em] -1(k + 1) &= 6k - 3 \\[0.6em] -k - 1 &= 6k - 3 \\[0.6em] -1 + 3 &= 6k + k \\[0.6em] 2 &= 7k \\[0.6em] k &= \dfrac{2}{7} \end{aligned}

Thus, the required ratio is $\dfrac{2}{7} : 1 = 2 : 7$ .

Step 2 · Verify with y-coordinate

Using the section formula for the $y$ -coordinate: $y = \dfrac{k y_2 + 1 \cdot y_1}{k + 1}$

Substitute $y = 6$ , $y_1 = 10$ , and $y_2 = -8$ :

\begin{aligned} 6 &= \dfrac{k(-8) + 1(10)}{k + 1} \\[0.6em] 6(k + 1) &= -8k + 10 \\[0.6em] 6k + 6 &= -8k + 10 \\[0.6em] 6k + 8k &= 10 - 6 \\[0.6em] 14k &= 4 \\[0.6em] k &= \dfrac{4}{14} \\[0.6em] k &= \dfrac{2}{7} \end{aligned}

Both coordinates yield $k = \dfrac{2}{7}$ .

Answer

$2 : 7$

Common Mistakes

Swapping Point Order: Taking $B$ first and $A$ second gives the ratio $7 : 2$ instead of $2 : 7$ . The ratio must correspond to the directed line segment from $A$ to $B$ .
Sign Error in Distribution: When expanding $-1(k + 1)$ , mistakenly writing $-k + 1$ instead of $-k - 1$ .
Skipping Verification: It is important to check the ratio in the $y$ -coordinate equation to confirm that the three points are indeed collinear.

More questions in Exercise 7.2

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

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 $1\text{ m}$ each. $100$ flower pots have been placed at a distance of $1\text{ m}$ from each other along $AD$ , as shown in Fig. 7.12. Niharika runs $\dfrac{1}{4}\text{th}$ the distance $AD$ on the 2nd line and posts a green flag. Preet runs $\dfrac{1}{5}\text{th}$ 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 segment 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.

If $(1, 2)$ , $(4, y)$ , $(x, 6)$ and $(3, 5)$ are the vertices of a parallelogram taken in order, find $x$ and $y$ .

Find the coordinates of a point $\text{A}$ , where $\text{AB}$ is the diameter of a circle whose centre is $(2, -3)$ and $\text{B}$ is $(1, 4)$ .

If A and B are $(-2, -2)$ and $(2, -4)$ , respectively, find the coordinates of P such that $\text{AP} = \dfrac{3}{7} \text{AB}$ and P lies on the line segment AB.

Find the coordinates of the points which divide the line segment joining $A(-2, 2)$ and $B(2, 8)$ into four equal parts.

Q10

Find the area of a rhombus if its vertices are $(3, 0)$ , $(4, 5)$ , $(-1, 4)$ and $(-2, -1)$ taken in order.

[Hint : Area of a rhombus = $\dfrac{1}{2}$ (product of its diagonals)]