Samacheer Kalvi Class 9 Maths Solutions Chapter 5 Coordinate Geometry Exercise 5.4

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.4

Question 1. Find the coordinates of the point which divides the line segment joining the points A (4,-3) and B (9,7) in the ratio 3 : 2.
Answer: We need to find a point P that divides the line segment connecting A(4, -3) and B(9, 7) in the ratio 3:2. This means \( m = 3 \) and \( n = 2 \). The coordinates of the points are \( x_1 = 4, y_1 = -3 \) and \( x_2 = 9, y_2 = 7 \). We use the section formula for internal division.
The point \( P(x, y) = \left( \frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n} \right) \)
Substitute the values:
\( x = \frac{(3)(9) + (2)(4)}{3+2} = \frac{27 + 8}{5} = \frac{35}{5} = 7 \)
\( y = \frac{(3)(7) + (2)(-3)}{3+2} = \frac{21 - 6}{5} = \frac{15}{5} = 3 \)
So, the coordinates of point P are \( (7, 3) \). This point is located on the segment AB.
In simple words: To find the point that splits a line in a certain way, we use a special formula. We plug in the numbers for the two end points and the given ratio, then do the math to get the coordinates of the dividing point.

🎯 Exam Tip: Remember the section formula for internal division: \( \left( \frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n} \right) \). Clearly label \( x_1, y_1, x_2, y_2, m, \) and \( n \) to avoid errors.

Question 2. In what ratio does the point P(2, -5) divide the line segment joining A(-3, 5) and B(4, -9).
Answer: We need to find the ratio \( m:n \) in which point P(2, -5) divides the line segment joining A(-3, 5) and B(4, -9).
Let the ratio be \( m:n \). The coordinates are \( x = 2, y = -5 \), \( x_1 = -3, y_1 = 5 \), \( x_2 = 4, y_2 = -9 \).
Using the section formula for the x-coordinate:
\( x = \frac{m x_2 + n x_1}{m+n} \)
\( 2 = \frac{m(4) + n(-3)}{m+n} \)
\( 2(m+n) = 4m - 3n \)
\( 2m + 2n = 4m - 3n \)
Now, we collect terms with \( m \) on one side and terms with \( n \) on the other:
\( 2n + 3n = 4m - 2m \)
\( 5n = 2m \)
\( \frac{m}{n} = \frac{5}{2} \)
So, the ratio \( m:n = 5:2 \). We can verify this with the y-coordinate as well:
\( y = \frac{m y_2 + n y_1}{m+n} \)
\( -5 = \frac{m(-9) + n(5)}{m+n} \)
\( -5(m+n) = -9m + 5n \)
\( -5m - 5n = -9m + 5n \)
\( -5m + 9m = 5n + 5n \)
\( 4m = 10n \)
\( \frac{m}{n} = \frac{10}{4} = \frac{5}{2} \)
Both coordinates give the same ratio, which confirms our answer. The point P divides the line segment in the ratio 5:2.
In simple words: We want to find how a point divides a line into two parts. We use the same formula, but this time we know the point's location and need to find the ratio. We set up an equation using the coordinates and solve for the ratio.

🎯 Exam Tip: When finding the ratio, you only need to use one coordinate (x or y). Using both can help cross-check your answer, but is not strictly necessary unless specified.

Question 3. Find the coordinates of a point P on the line segment joining A(1, 2) and B(6, 7) in such a way that AP = \( \frac{2}{5} \) AB.
Answer: We are given that point P lies on the line segment AB such that \( AP = \frac{2}{5} AB \).
This means that if the total length of AB is 5 parts, then AP is 2 parts. This also implies that PB is \( AB - AP = 5 - 2 = 3 \) parts. Therefore, the ratio \( AP:PB \) is \( 2:3 \).
So, \( m = 2 \) and \( n = 3 \). The coordinates of the points are \( x_1 = 1, y_1 = 2 \) and \( x_2 = 6, y_2 = 7 \).
Using the section formula for internal division:
The point \( P(x, y) = \left( \frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n} \right) \)
Substitute the values:
\( x = \frac{(2)(6) + (3)(1)}{2+3} = \frac{12 + 3}{5} = \frac{15}{5} = 3 \)
\( y = \frac{(2)(7) + (3)(2)}{2+3} = \frac{14 + 6}{5} = \frac{20}{5} = 4 \)
So, the coordinates of point P are \( (3, 4) \). This point correctly divides the segment as required.
In simple words: We are told how one part of a line relates to the whole line. From this, we figure out the ratio in which the point divides the line. Then, we use the section formula with that ratio and the end points to find the point's exact location.

🎯 Exam Tip: When given a relationship like \( AP = \frac{k}{L} AB \), make sure to correctly deduce the ratio \( AP:PB \). If \( AP = \frac{2}{5} AB \), then \( AP:PB \) is \( 2:(5-2) \), or \( 2:3 \).

Question 4. Find the coordinates of the points of trisection of the line segment joining the points A (-5, 6) and B (4, -3).
Answer: Trisection means dividing a line segment into three equal parts. For a segment AB, there will be two points, P and Q, such that \( AP = PQ = QB \).
This implies that point P divides AB in the ratio \( 1:2 \), and point Q divides AB in the ratio \( 2:1 \).
The coordinates of the end points are \( A(x_1, y_1) = (-5, 6) \) and \( B(x_2, y_2) = (4, -3) \).

For Point P (dividing in ratio 1:2): Here, \( m=1, n=2 \).
\( x_P = \frac{m x_2 + n x_1}{m+n} = \frac{(1)(4) + (2)(-5)}{1+2} = \frac{4 - 10}{3} = \frac{-6}{3} = -2 \)
\( y_P = \frac{m y_2 + n y_1}{m+n} = \frac{(1)(-3) + (2)(6)}{1+2} = \frac{-3 + 12}{3} = \frac{9}{3} = 3 \)
So, the coordinates of Point P are \( (-2, 3) \).

For Point Q (dividing in ratio 2:1): Here, \( m=2, n=1 \).
\( x_Q = \frac{m x_2 + n x_1}{m+n} = \frac{(2)(4) + (1)(-5)}{2+1} = \frac{8 - 5}{3} = \frac{3}{3} = 1 \)
\( y_Q = \frac{m y_2 + n y_1}{m+n} = \frac{(2)(-3) + (1)(6)}{2+1} = \frac{-6 + 6}{3} = \frac{0}{3} = 0 \)
So, the coordinates of Point Q are \( (1, 0) \).
The points of trisection are P(-2, 3) and Q(1, 0). These two points break the line segment into three equal pieces.
In simple words: To split a line into three equal parts, we find two special points. One point is found by using a 1:2 ratio, and the other by using a 2:1 ratio. We apply the section formula for each ratio with the given end points.

🎯 Exam Tip: Always remember that for trisection, there are two points, and they divide the segment in ratios 1:2 and 2:1, respectively. Keep track of which point corresponds to which ratio.

Question 5. The line segment joining A(6, 3) and B(-1, -4) is doubled in length by adding half of AB to each end. Find the coordinates of the new end points.
Answer: Let the original line segment be AB with A(6, 3) and B(-1, -4).
First, we find the length of AB using the distance formula:
\( AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(-1 - 6)^2 + (-4 - 3)^2} \)
\( AB = \sqrt{(-7)^2 + (-7)^2} = \sqrt{49 + 49} = \sqrt{98} = \sqrt{49 \times 2} = 7\sqrt{2} \)
The problem states that half of AB is added to each end. This means the new total length will be \( AB + \frac{1}{2}AB + \frac{1}{2}AB = 2AB \). Let the new endpoints be A' and B'. The diagram of the points would be B' - A - B - A'.
Therefore, A is the midpoint of B'B, and B is the midpoint of AA'. This is incorrect. The problem implies B' is on the extension of BA, and A' is on the extension of AB. So, B' - A - B - A' is the correct order.
In this setup, \( B'A = \frac{1}{2} AB \) and \( AA' = \frac{1}{2} AB \).
This means A divides the segment B'B externally in the ratio \( \frac{AB'}{AB} = \frac{B'A + AB}{AB} = \frac{\frac{1}{2}AB + AB}{AB} = \frac{\frac{3}{2}AB}{AB} = \frac{3}{2} \).
So, A divides B'B externally in the ratio 3:2. (This is one way to think of it, but the simpler interpretation below is better.)

A more straightforward approach is to realize that A is between B' and B, and B is between A and A'.
Thus, A is the midpoint of B'B. This is also not fully correct based on problem wording. "Adding half of AB to each end" means extending the segment beyond A and beyond B by AB/2.
So, if A is (6,3) and B is (-1,-4):
Let A' be the new endpoint on the side of B. This means B is between A and A', and \( BA' = \frac{1}{2} AB \).
So, A' divides AB externally in the ratio \( (AB + BA'):BA' = (AB + \frac{1}{2}AB):\frac{1}{2}AB = \frac{3}{2}AB:\frac{1}{2}AB = 3:1 \).
To find A', we use the external section formula with A(6,3) as \( x_1, y_1 \) and B(-1,-4) as \( x_2, y_2 \), and the ratio \( m:n = 3:1 \).
\( x_{A'} = \frac{m x_2 - n x_1}{m-n} = \frac{3(-1) - 1(6)}{3-1} = \frac{-3 - 6}{2} = \frac{-9}{2} \)
\( y_{A'} = \frac{m y_2 - n y_1}{m-n} = \frac{3(-4) - 1(3)}{3-1} = \frac{-12 - 3}{2} = \frac{-15}{2} \)
So, one new endpoint, A', is \( \left(-\frac{9}{2}, -\frac{15}{2}\right) \).

Let B' be the new endpoint on the side of A. This means A is between B' and B, and \( B'A = \frac{1}{2} AB \).
So, B divides B'A externally in the ratio \( (B'A + AB):B'A = (\frac{1}{2}AB + AB):\frac{1}{2}AB = \frac{3}{2}AB:\frac{1}{2}AB = 3:1 \).
To find B', we use the external section formula with B(-1,-4) as \( x_1, y_1 \) and A(6,3) as \( x_2, y_2 \), and the ratio \( m:n = 3:1 \).
\( x_{B'} = \frac{m x_2 - n x_1}{m-n} = \frac{3(6) - 1(-1)}{3-1} = \frac{18 + 1}{2} = \frac{19}{2} \)
\( y_{B'} = \frac{m y_2 - n y_1}{m-n} = \frac{3(3) - 1(-4)}{3-1} = \frac{9 + 4}{2} = \frac{13}{2} \)
So, the other new endpoint, B', is \( \left(\frac{19}{2}, \frac{13}{2}\right) \).
The coordinates of the new end points are \( \left(-\frac{9}{2}, -\frac{15}{2}\right) \) and \( \left(\frac{19}{2}, \frac{13}{2}\right) \). These points are found by extending the original line segment on both sides.
In simple words: First, we find the length of the original line. Then, we imagine extending the line by half its length on both sides. To find these new end points, we use a formula for "external division" because the new points are outside the original line segment. We apply this formula twice, once for each side of the original line.

🎯 Exam Tip: "Adding half of AB to each end" means the total length becomes \( 2AB \). When using the section formula for external division, ensure you correctly identify the 'outside' point and the ratio for the extension.

Question 6. Using section formula, show that the points A (7, -5), B (9, -3) and C (13, 1) are collinear.
Answer: To show that three points are collinear using the section formula, we need to show that one of the points divides the line segment joining the other two in some ratio.
Let's assume point B divides the line segment AC in a ratio \( m:n \). The coordinates are \( A(x_1, y_1) = (7, -5) \), \( C(x_2, y_2) = (13, 1) \), and the dividing point \( B(x, y) = (9, -3) \).
Using the section formula for the x-coordinate of B:
\( 9 = \frac{m(13) + n(7)}{m+n} \)
\( 9(m+n) = 13m + 7n \)
\( 9m + 9n = 13m + 7n \)
\( 9n - 7n = 13m - 9m \)
\( 2n = 4m \)
\( \frac{m}{n} = \frac{2}{4} = \frac{1}{2} \)
So, the ratio is \( 1:2 \). Let's check this with the y-coordinate of B:
\( -3 = \frac{m(1) + n(-5)}{m+n} \)
\( -3(m+n) = m - 5n \)
\( -3m - 3n = m - 5n \)
\( -3n + 5n = m + 3m \)
\( 2n = 4m \)
\( \frac{m}{n} = \frac{2}{4} = \frac{1}{2} \)
Since both the x and y coordinates yield the same ratio \( 1:2 \), it means that point B divides the line segment AC in the ratio \( 1:2 \). Therefore, the three points A, B, and C are collinear. If B was not on the line AC, a consistent ratio would not be found.
In simple words: We check if the middle point divides the line segment formed by the other two points in a consistent ratio. If it does, using the section formula, then all three points lie on the same straight line.

🎯 Exam Tip: To prove collinearity using the section formula, choose one point as the divisor and the other two as the endpoints. If the ratio \( m:n \) is consistent for both x and y coordinates, the points are collinear.

Question 7. A line segment AB is increased along its length by 25% by producing it to C on the side of B. If A and B have the coordinates (-2, -3) and (2, 1) respectively, then find the coordinates of C.
Answer: We are given the coordinates of A(-2, -3) and B(2, 1). The line segment AB is extended to C such that C is on the side of B.
The length of the segment is increased by 25%. This means the new length AC is \( AB + 25\% \text{ of } AB = AB + \frac{1}{4}AB = \frac{5}{4}AB \).
Since C is on the side of B (A-B-C), we have \( AC = AB + BC \).
So, \( \frac{5}{4}AB = AB + BC \)
\( BC = \frac{5}{4}AB - AB = \frac{1}{4}AB \).
This implies that the ratio \( AB:BC = 4:1 \).
Since B lies between A and C, point B divides the segment AC internally in the ratio \( 4:1 \).
Let the coordinates of C be \( (x_C, y_C) \). We have A(\( x_1, y_1 \)) = (-2, -3) and C(\( x_2, y_2 \)) = (\( x_C, y_C \)). The dividing point is B(\( x, y \)) = (2, 1). The ratio is \( m:n = 4:1 \).
Using the section formula for internal division:
\( x = \frac{m x_2 + n x_1}{m+n} \)
\( 2 = \frac{4(x_C) + 1(-2)}{4+1} \)
\( 2 = \frac{4x_C - 2}{5} \)
\( 10 = 4x_C - 2 \)
\( 12 = 4x_C \)
\( x_C = 3 \)

Now for the y-coordinate:
\( y = \frac{m y_2 + n y_1}{m+n} \)
\( 1 = \frac{4(y_C) + 1(-3)}{4+1} \)
\( 1 = \frac{4y_C - 3}{5} \)
\( 5 = 4y_C - 3 \)
\( 8 = 4y_C \)
\( y_C = 2 \)
Therefore, the coordinates of C are \( (3, 2) \). This point represents the extended end of the line segment.
In simple words: We are told a line is stretched beyond one of its ends. First, we figure out how much the original line's length relates to the added part. This helps us find the ratio in which the original end point divides the new, longer segment. Then we use the section formula to find the coordinates of the new, extended point.

🎯 Exam Tip: Pay close attention to whether a point divides a segment internally or externally. In this case, since C extends beyond B, B divides AC internally.