OP Malhotra Class 11 Maths Solutions Chapter 15 Basic Concepts of Points and their Coordinates Exercise 15 (B)

Question 1. Find the mid-points of the lines joining
(i) (5, 8),(9, 11);
(ii) (0, 0),(8, -5);
(iii) (-7, 0),(0, 10);
(iv) (-4, 3) and (6, -7)
Answer: We use the mid-point formula for a line segment joining points \( (x_1, y_1) \) and \( (x_2, y_2) \), which is \( \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \). This formula helps us find the exact center of any line.
(i) For points (5, 8) and (9, 11):
Mid-point \( = \left(\frac{5+9}{2}, \frac{8+11}{2}\right) = \left(\frac{14}{2}, \frac{19}{2}\right) = \left(7, \frac{19}{2}\right) \)
(ii) For points (0, 0) and (8, -5):
Mid-point \( = \left(\frac{0+8}{2}, \frac{0-5}{2}\right) = \left(\frac{8}{2}, \frac{-5}{2}\right) = \left(4, \frac{-5}{2}\right) \)
(iii) For points (-7, 0) and (0, 10):
Mid-point \( = \left(\frac{-7+0}{2}, \frac{0+10}{2}\right) = \left(\frac{-7}{2}, \frac{10}{2}\right) = \left(\frac{-7}{2}, 5\right) \)
(iv) For points (-4, 3) and (6, -7):
Mid-point \( = \left(\frac{-4+6}{2}, \frac{3-7}{2}\right) = \left(\frac{2}{2}, \frac{-4}{2}\right) = (1, -2) \)
In simple words: To find the middle of a line, you add the x-coordinates and divide by two, then do the same for the y-coordinates. This gives you the coordinates of the point that is exactly halfway between the two given points.

🎯 Exam Tip: Remember the mid-point formula: \( \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \). Double-check your arithmetic, especially with negative numbers, to avoid calculation errors.

Question 2. Find the mid-points of the sides of a triangle whose vertices are A(1, -1), B (4, -1), C (4, 3).
Answer: We need to find the mid-point of each side of the triangle.
For side BC, using B(4, -1) and C(4, 3):
Mid-point of BC \( = \left(\frac{4+4}{2}, \frac{-1+3}{2}\right) = \left(\frac{8}{2}, \frac{2}{2}\right) = (4, 1) \)
For side CA, using C(4, 3) and A(1, -1):
Mid-point of CA \( = \left(\frac{1+4}{2}, \frac{-1+3}{2}\right) = \left(\frac{5}{2}, \frac{2}{2}\right) = \left(\frac{5}{2}, 1\right) \)
For side AB, using A(1, -1) and B(4, -1):
Mid-point of AB \( = \left(\frac{1+4}{2}, \frac{-1-1}{2}\right) = \left(\frac{5}{2}, \frac{-2}{2}\right) = \left(\frac{5}{2}, -1\right) \)
These three midpoints form a new triangle inside the original one.
The triangle with its midpoints is shown below:

In simple words: A triangle has three sides. For each side, we find the middle point using the mid-point formula. This gives us three new points, each exactly in the middle of one of the triangle's sides.

🎯 Exam Tip: Always label your vertices (A, B, C) clearly before applying the mid-point formula to each pair of vertices that form a side.

Question 3. Find the centre of a circle if the end points of a diameter are A(-5, 7) and B (3, -11).
Answer: The center of a circle is always the mid-point of its diameter. To find the center, we use the mid-point formula for the two given endpoints of the diameter.
Given endpoints of the diameter are A(-5, 7) and B(3, -11).
Let the center of the circle be C.
Coordinates of center C \( = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \)
\( = \left(\frac{-5+3}{2}, \frac{7-11}{2}\right) \)
\( = \left(\frac{-2}{2}, \frac{-4}{2}\right) \)
\( = (-1, -2) \)
So, the coordinates of the center C are (-1, -2). This is a fundamental property of circles.
The diagram shows the diameter AB and its center C:

In simple words: The center of a circle is always exactly in the middle of any line that goes through it and touches both sides, called a diameter. So, you just find the mid-point of the diameter's ends.

🎯 Exam Tip: Understand that the center of a circle is equidistant from all points on its circumference, and a diameter passes through the center. This makes the center the midpoint of any diameter.

Question 4. If M is the mid-point of AB, find the coordinates of :
(i) A if the co-ordinates of M and B are M(2, 8) and B(-4, 19) and
(ii) B if the co-ordinates of A and M are A(-1, 2), M(-2, 4)
Answer: We use the mid-point formula \( \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \) and rearrange it to find the missing coordinate.
(i) Given M(2, 8) and B(-4, 19). Let A be \( (\alpha, \beta) \).
Since M is the mid-point of AB:
\( (2, 8) = \left(\frac{-4+\alpha}{2}, \frac{19+\beta}{2}\right) \)
Comparing the x-coordinates:
\( 2 = \frac{-4+\alpha}{2} \)
\( \implies 4 = -4+\alpha \)
\( \implies \alpha = 8 \)
Comparing the y-coordinates:
\( 8 = \frac{19+\beta}{2} \)
\( \implies 16 = 19+\beta \)
\( \implies \beta = -3 \)
So, the coordinates of A are (8, -3).
(ii) Given A(-1, 2) and M(-2, 4). Let B be \( (\alpha, \beta) \).
Since M is the mid-point of AB:
\( (-2, 4) = \left(\frac{-1+\alpha}{2}, \frac{2+\beta}{2}\right) \)
Comparing the x-coordinates:
\( -2 = \frac{-1+\alpha}{2} \)
\( \implies -4 = -1+\alpha \)
\( \implies \alpha = -3 \)
Comparing the y-coordinates:
\( 4 = \frac{2+\beta}{2} \)
\( \implies 8 = 2+\beta \)
\( \implies \beta = 6 \)
So, the coordinates of B are (-3, 6).
In simple words: If you know the middle point and one end of a line, you can find the other end. You just double the middle point's coordinates and subtract the known end point's coordinates. This helps us find the missing coordinate.

🎯 Exam Tip: When finding a missing endpoint, remember to set up two separate equations (one for x and one for y) using the mid-point formula, and solve them individually. This systematic approach reduces errors.

Question 5. Find the distance between each of the following pairs of points :
(i) (7, 9),(4, 5);
(ii) (15, 11),(3, 6);
(iii) (4, -5), (0, 0);
(iv) (2, -11), (-4, -3)
Answer: We use the distance formula \( d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \). This formula comes from the Pythagorean theorem and helps us calculate the length of a line segment.
(i) Distance between A(7, 9) and B(4, 5):
\( |AB| = \sqrt{(4-7)^2 + (5-9)^2} \)
\( = \sqrt{(-3)^2 + (-4)^2} \)
\( = \sqrt{9+16} = \sqrt{25} = 5 \) units.
(ii) Distance between C(15, 11) and D(3, 6):
\( |CD| = \sqrt{(3-15)^2 + (6-11)^2} \)
\( = \sqrt{(-12)^2 + (-5)^2} \)
\( = \sqrt{144+25} = \sqrt{169} = 13 \) units.
(iii) Distance between L(4, -5) and M(0, 0):
\( |LM| = \sqrt{(0-4)^2 + (0-(-5))^2} \)
\( = \sqrt{(-4)^2 + (5)^2} \)
\( = \sqrt{16+25} = \sqrt{41} \) units.
(iv) Distance between P(2, -11) and Q(-4, -3):
\( |PQ| = \sqrt{(-4-2)^2 + (-3-(-11))^2} \)
\( = \sqrt{(-6)^2 + (8)^2} \)
\( = \sqrt{36+64} = \sqrt{100} = 10 \) units.
In simple words: To find how far apart two points are, you can imagine a right triangle between them. You square the difference in their x-values, square the difference in their y-values, add those two squared numbers, and then find the square root of the total. This gives you the straight-line distance.

🎯 Exam Tip: Be careful with signs when subtracting coordinates, especially with negative numbers. Squaring a negative number always results in a positive number, which is a common point of error.

Question 6. Find the radius of the circle that has its centre at (0, -4) and passes through \( (\sqrt{13}, 2) \)
Answer: The radius of a circle is the distance from its center to any point on its circumference. We can find this distance using the distance formula.
Given center C(0, -4) and a point A\( (\sqrt{13}, 2) \) on the circle.
The radius \( r \) is the distance |CA|.
\( r = \sqrt{(\sqrt{13}-0)^2 + (2-(-4))^2} \)
\( = \sqrt{(\sqrt{13})^2 + (2+4)^2} \)
\( = \sqrt{13 + 6^2} \)
\( = \sqrt{13 + 36} \)
\( = \sqrt{49} = 7 \) units.
The radius of the circle is 7 units. This means any point on the circle is 7 units away from the center.
The circle with its center and a point on its circumference is shown below:

In simple words: The radius is simply the distance from the center of the circle to any point on its edge. We calculate this distance using the distance formula.

🎯 Exam Tip: Understand that the distance formula is key for finding lengths in coordinate geometry. Here, it directly applies to finding the radius of a circle given its center and a point on its edge.

Question 7. Find the lengths of the sides of the triangle whose vertices are A(3, 4), B(2, -1) and C(4, -6).
Answer: To find the length of each side of the triangle, we will use the distance formula between the two vertices that form each side. The distance formula is \( d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \).
Length of side AB, using A(3, 4) and B(2, -1):
\( |AB| = \sqrt{(2-3)^2 + (-1-4)^2} \)
\( = \sqrt{(-1)^2 + (-5)^2} \)
\( = \sqrt{1+25} = \sqrt{26} \) units.
Length of side BC, using B(2, -1) and C(4, -6):
\( |BC| = \sqrt{(4-2)^2 + (-6-(-1))^2} \)
\( = \sqrt{(2)^2 + (-5)^2} \)
\( = \sqrt{4+25} = \sqrt{29} \) units.
Length of side AC, using A(3, 4) and C(4, -6):
\( |AC| = \sqrt{(4-3)^2 + (-6-4)^2} \)
\( = \sqrt{(1)^2 + (-10)^2} \)
\( = \sqrt{1+100} = \sqrt{101} \) units.
Each side length is a positive value, as distance cannot be negative.
The triangle with its vertices is shown below:

In simple words: To find the length of each side of a triangle, we calculate the distance between its corner points. We do this three times, once for each pair of corners that form a side.

🎯 Exam Tip: Always write down the distance formula first. Be careful with calculations involving negative numbers and ensure you square both the x and y differences before adding them.

Question 8. The vertices of △ABC are A(-1, 3), B(1, 1) and C(5, 1). Find the length of the median to (i) AB, (ii) AC, (iii) BC.
Answer: A median of a triangle connects a vertex to the mid-point of the opposite side. First, we'll find the mid-point of each side, and then calculate the distance from the opposite vertex to that mid-point.
(i) Median to side AB:
Let D be the mid-point of AB. Vertices A(-1, 3) and B(1, 1).
Coordinates of D \( = \left(\frac{-1+1}{2}, \frac{3+1}{2}\right) = \left(\frac{0}{2}, \frac{4}{2}\right) = (0, 2) \)
The median to AB is CD. We find the length of CD using C(5, 1) and D(0, 2).
Length of median |CD| \( = \sqrt{(0-5)^2 + (2-1)^2} \)
\( = \sqrt{(-5)^2 + (1)^2} = \sqrt{25+1} = \sqrt{26} \) units.
(ii) Median to side AC:
Let E be the mid-point of AC. Vertices A(-1, 3) and C(5, 1).
Coordinates of E \( = \left(\frac{-1+5}{2}, \frac{3+1}{2}\right) = \left(\frac{4}{2}, \frac{4}{2}\right) = (2, 2) \)
The median to AC is BE. We find the length of BE using B(1, 1) and E(2, 2).
Length of median |BE| \( = \sqrt{(2-1)^2 + (2-1)^2} \)
\( = \sqrt{(1)^2 + (1)^2} = \sqrt{1+1} = \sqrt{2} \) units.
(iii) Median to side BC:
Let F be the mid-point of BC. Vertices B(1, 1) and C(5, 1).
Coordinates of F \( = \left(\frac{1+5}{2}, \frac{1+1}{2}\right) = \left(\frac{6}{2}, \frac{2}{2}\right) = (3, 1) \)
The median to BC is AF. We find the length of AF using A(-1, 3) and F(3, 1).
Length of median |AF| \( = \sqrt{(3-(-1))^2 + (1-3)^2} \)
\( = \sqrt{(4)^2 + (-2)^2} = \sqrt{16+4} = \sqrt{20} = 2\sqrt{5} \) units.
Medians connect vertices to the midpoints of opposite sides, adding structure to the triangle.
The triangle with its medians is shown below:

In simple words: A median of a triangle connects a corner to the middle of the side opposite it. First, find the middle point of the side, then calculate the distance from the opposite corner to that middle point.

🎯 Exam Tip: When finding medians, correctly identify the mid-point of the side and the opposite vertex. Any mistake in calculating the mid-point will lead to an incorrect median length.

Question 9. A circle has its centre at the origin and a radius of \( \sqrt{12} \). State whether each of the following points is on, outside or inside the circle : \( (1, -\sqrt{7}),(3, 5),(2, 2\sqrt{2}) \).
Answer: The center of the circle is at the origin (0, 0), and its radius \( r = \sqrt{12} \). To determine if a point is inside, outside, or on the circle, we calculate its distance from the origin (center) and compare it to the radius.
Let the given points be P\( (1, -\sqrt{7}) \), Q(3, 5), and R\( (2, 2\sqrt{2}) \).
1. For point P\( (1, -\sqrt{7}) \):
Distance from origin \( |CP| = \sqrt{(1-0)^2 + (-\sqrt{7}-0)^2} \)
\( = \sqrt{1^2 + (-\sqrt{7})^2} = \sqrt{1+7} = \sqrt{8} \)
Since \( \sqrt{8} < \sqrt{12} \) (because \( 8 < 12 \)), point P lies **inside** the circle.
2. For point Q(3, 5):
Distance from origin \( |CQ| = \sqrt{(3-0)^2 + (5-0)^2} \)
\( = \sqrt{3^2 + 5^2} = \sqrt{9+25} = \sqrt{34} \)
Since \( \sqrt{34} > \sqrt{12} \) (because \( 34 > 12 \)), point Q lies **outside** the circle.
3. For point R\( (2, 2\sqrt{2}) \):
Distance from origin \( |CR| = \sqrt{(2-0)^2 + (2\sqrt{2}-0)^2} \)
\( = \sqrt{2^2 + (2\sqrt{2})^2} = \sqrt{4+ (4 \times 2)} = \sqrt{4+8} = \sqrt{12} \)
Since \( \sqrt{12} = r \), point R lies **on** the circle.
The distance from the center tells us its position relative to the circle's edge.
The circle with its center at the origin is shown below:

In simple words: A point is inside the circle if its distance from the center is less than the radius. It's on the circle if the distance equals the radius, and outside if the distance is greater than the radius.

🎯 Exam Tip: The key is to calculate the distance from the point to the circle's center using the distance formula. Then, compare this distance directly to the given radius \( r \). Squaring both sides (\( d^2 \) and \( r^2 \)) can sometimes simplify comparisons if square roots are complex.

Question 10. A line is of length 10, and one end is at the point (-3, 2). If the ordinate of the other end be 10, prove that the abscissa will be 3 or -9.
Answer: Let the known end of the line be P(-3, 2). Let the other end be Q(x, 10), where x is the unknown abscissa. The length of the line PQ is given as 10 units. We will use the distance formula to find x.
Distance \( |PQ| = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \)
Given \( |PQ| = 10 \), P(-3, 2), and Q(x, 10).
\( 10 = \sqrt{(x-(-3))^2 + (10-2)^2} \)
\( 10 = \sqrt{(x+3)^2 + (8)^2} \)
\( 10 = \sqrt{(x+3)^2 + 64} \)
Square both sides to remove the square root:
\( 10^2 = (x+3)^2 + 64 \)
\( 100 = (x+3)^2 + 64 \)
Subtract 64 from both sides:
\( (x+3)^2 = 100 - 64 \)
\( (x+3)^2 = 36 \)
Take the square root of both sides:
\( x+3 = \pm \sqrt{36} \)
\( x+3 = \pm 6 \)
This gives two possibilities:
Case 1: \( x+3 = 6 \)
\( \implies x = 6-3 \)
\( \implies x = 3 \)
Case 2: \( x+3 = -6 \)
\( \implies x = -6-3 \)
\( \implies x = -9 \)
Therefore, the abscissa (x-coordinate) of the other end can be 3 or -9. This demonstrates that there can be multiple points at a specific distance and y-coordinate.
In simple words: We know one point, the length of the line, and the y-value of the other point. By using the distance formula and some algebra, we can find the two possible x-values for the second point.

🎯 Exam Tip: When solving equations involving squares, remember to consider both the positive and negative square roots. This is crucial for finding all possible solutions, as seen in this problem.

Question 11. Find the coordinates of the point which divides internally the join of the points
(i) (8, 9) and (-7, 4) in the ratio 2 : 3;
(ii) (1, -2) and (4, 7) in the ratio 1 : 2.
Answer: We use the section formula for internal division: if a point P(x, y) divides the line segment joining \( (x_1, y_1) \) and \( (x_2, y_2) \) in the ratio m:n, then \( P(x,y) = \left(\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}\right) \). This formula helps us locate points that split a line segment into specific proportions.
(i) Points A(8, 9) and B(-7, 4), ratio 2:3 (m=2, n=3).
Let P(x, y) be the dividing point.
\( P(x,y) = \left(\frac{2 \times (-7) + 3 \times 8}{2+3}, \frac{2 \times 4 + 3 \times 9}{2+3}\right) \)
\( = \left(\frac{-14+24}{5}, \frac{8+27}{5}\right) \)
\( = \left(\frac{10}{5}, \frac{35}{5}\right) \)
\( = (2, 7) \)
The point P divides the line segment AB internally in the ratio 2:3.
The line segment AB with point P is shown below:

(ii) Points A(1, -2) and B(4, 7), ratio 1:2 (m=1, n=2).
Let Q(x, y) be the dividing point.
\( Q(x,y) = \left(\frac{1 \times 4 + 2 \times 1}{1+2}, \frac{1 \times 7 + 2 \times (-2)}{1+2}\right) \)
\( = \left(\frac{4+2}{3}, \frac{7-4}{3}\right) \)
\( = \left(\frac{6}{3}, \frac{3}{3}\right) \)
\( = (2, 1) \)
The point Q divides the line segment AB internally in the ratio 1:2.
In simple words: To find a point that divides a line segment into a certain ratio from the inside, you use a special formula. You multiply each coordinate of the end points by the opposite ratio number, add them up, and then divide by the sum of the ratios.

🎯 Exam Tip: Pay close attention to which point corresponds to \( (x_1, y_1) \) and which to \( (x_2, y_2) \), and the order of m:n in the ratio. Swapping these can lead to incorrect results.

Question 12. Find the coordinates of the point which divides externally the join of the points
(i) (-4, 4) and (1, 7) in the ratio 2 : 1;
(ii) (3, 4) and (-6, 2) in the ratio 3 : 2.
Answer: We use the section formula for external division. If a point P(x, y) divides the line segment joining \( (x_1, y_1) \) and \( (x_2, y_2) \) externally in the ratio m:n, then \( P(x,y) = \left(\frac{mx_2-nx_1}{m-n}, \frac{my_2-ny_1}{m-n}\right) \). This formula helps us find a point that lies *outside* the line segment, extending it in a specific ratio. An external division with ratio m:n is equivalent to an internal division with ratio m:(-n).
(i) Points A(-4, 4) and B(1, 7) in the ratio 2:1 (m=2, n=1).
Using the external division formula:
\( P(x,y) = \left(\frac{2 \times 1 - 1 \times (-4)}{2-1}, \frac{2 \times 7 - 1 \times 4}{2-1}\right) \)
\( = \left(\frac{2+4}{1}, \frac{14-4}{1}\right) \)
\( = (6, 10) \)
Alternatively, using internal division with ratio 2:(-1):
\( P(x,y) = \left(\frac{2 \times 1 + (-1) \times (-4)}{2+(-1)}, \frac{2 \times 7 + (-1) \times 4}{2+(-1)}\right) \)
\( = \left(\frac{2+4}{1}, \frac{14-4}{1}\right) \)
\( = (6, 10) \)
Both methods yield the same result. The point (6, 10) is located outside the segment AB.
The line segment AB with point P is shown below:

(ii) Points A(3, 4) and B(-6, 2) in the ratio 3:2 (m=3, n=2).
Using the external division formula:
\( P(x,y) = \left(\frac{3 \times (-6) - 2 \times 3}{3-2}, \frac{3 \times 2 - 2 \times 4}{3-2}\right) \)
\( = \left(\frac{-18-6}{1}, \frac{6-8}{1}\right) \)
\( = (-24, -2) \)
Alternatively, using internal division with ratio 3:(-2):
\( P(x,y) = \left(\frac{3 \times (-6) + (-2) \times 3}{3+(-2)}, \frac{3 \times 2 + (-2) \times 4}{3+(-2)}\right) \)
\( = \left(\frac{-18-6}{1}, \frac{6-8}{1}\right) \)
\( = (-24, -2) \)
The point (-24, -2) is located outside the segment AB.
In simple words: When a point divides a line externally, it means the point lies on the extended line, not between the two given points. The formula is similar to internal division but uses subtraction in both the numerator and denominator. You can also think of it as an internal division with a negative ratio.

🎯 Exam Tip: The key difference in the external section formula is the minus sign: \( \left(\frac{mx_2-nx_1}{m-n}, \frac{my_2-ny_1}{m-n}\right) \). Be very careful with the signs, especially when one of the coordinates is negative.

Question 13. Find the coordinates of the points of trisection of the line joining the points (2, 3) and (6, 5).
Answer: Trisection means dividing a line segment into three equal parts. If a line segment AB is trisected by points P and Q, then P divides AB in the ratio 1:2 internally, and Q divides AB in the ratio 2:1 internally. We will use the section formula for internal division: \( P(x,y) = \left(\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}\right) \).
Let the given points be A(2, 3) and B(6, 5).
1. Find point P (divides AB in ratio 1:2):
Here, m=1 and n=2.
\( P(x,y) = \left(\frac{1 \times 6 + 2 \times 2}{1+2}, \frac{1 \times 5 + 2 \times 3}{1+2}\right) \)
\( = \left(\frac{6+4}{3}, \frac{5+6}{3}\right) \)
\( = \left(\frac{10}{3}, \frac{11}{3}\right) \)
2. Find point Q (divides AB in ratio 2:1):
Here, m=2 and n=1.
\( Q(x,y) = \left(\frac{2 \times 6 + 1 \times 2}{2+1}, \frac{2 \times 5 + 1 \times 3}{2+1}\right) \)
\( = \left(\frac{12+2}{3}, \frac{10+3}{3}\right) \)
\( = \left(\frac{14}{3}, \frac{13}{3}\right) \)
The coordinates of the points of trisection are \( \left(\frac{10}{3}, \frac{11}{3}\right) \) and \( \left(\frac{14}{3}, \frac{13}{3}\right) \). These two points divide the segment AB into three equal length portions.
The line segment AB with points of trisection P and Q is shown below:

In simple words: To split a line into three equal parts, you find two points. The first point divides the line in a 1:2 ratio, and the second point divides it in a 2:1 ratio. You use the section formula for both.

🎯 Exam Tip: When trisection is asked, remember you need to find two points. It's easy to forget one or mix up the ratios (1:2 and 2:1). Also, ensure you apply the internal section formula correctly.

Question 14. The line joining the points (3, 2) and (6, 8) is divided into four equal parts, find the coordinates of the points of section.
Answer: If a line segment AB is divided into four equal parts, there will be three points of section, say P, Q, and R.
P divides AB in the ratio 1:3 internally.
Q divides AB in the ratio 2:2 (i.e., 1:1) internally, meaning Q is the mid-point of AB.
R divides AB in the ratio 3:1 internally.
We will use the internal section formula: \( P(x,y) = \left(\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}\right) \).
Let the given points be A(3, 2) and B(6, 8).
1. Find point P (ratio 1:3): (m=1, n=3)
\( P(x,y) = \left(\frac{1 \times 6 + 3 \times 3}{1+3}, \frac{1 \times 8 + 3 \times 2}{1+3}\right) \)
\( = \left(\frac{6+9}{4}, \frac{8+6}{4}\right) \)
\( = \left(\frac{15}{4}, \frac{14}{4}\right) = \left(\frac{15}{4}, \frac{7}{2}\right) \)
2. Find point Q (ratio 1:1, or mid-point of AB): (m=1, n=1)
\( Q(x,y) = \left(\frac{3+6}{2}, \frac{2+8}{2}\right) \)
\( = \left(\frac{9}{2}, \frac{10}{2}\right) = \left(\frac{9}{2}, 5\right) \)
3. Find point R (ratio 3:1): (m=3, n=1)
\( R(x,y) = \left(\frac{3 \times 6 + 1 \times 3}{3+1}, \frac{3 \times 8 + 1 \times 2}{3+1}\right) \)
\( = \left(\frac{18+3}{4}, \frac{24+2}{4}\right) \)
\( = \left(\frac{21}{4}, \frac{26}{4}\right) = \left(\frac{21}{4}, \frac{13}{2}\right) \)
The coordinates of the points of section are \( \left(\frac{15}{4}, \frac{7}{2}\right), \left(\frac{9}{2}, 5\right), \left(\frac{21}{4}, \frac{13}{2}\right) \). These three points precisely divide the line segment into four equal parts.
The line segment AB with points of section P, Q, and R is shown below:

In simple words: To divide a line into four equal parts, you need three points. The first point is at a 1:3 ratio, the second is at the midpoint (1:1), and the third is at a 3:1 ratio. Each of these points can be found using the section formula.

🎯 Exam Tip: For dividing a line into N equal parts, there will be N-1 points. Identify the correct ratios for each point (e.g., for 4 parts, points are at 1:3, 1:1, 3:1). Always simplify the ratios before applying the formula.

Question 15. In what ratio does the point \( \left(1, \frac{-7}{2}\right) \) divide the join of (-2, -4) and \( \left(2, \frac{-10}{3}\right) \)?
Answer: Let the given points be A(-2, -4) and B\( \left(2, \frac{-10}{3}\right) \). Let the dividing point be P\( \left(1, \frac{-7}{2}\right) \). We assume P divides AB internally in the ratio k:1. If k turns out to be negative, it means the division is external. We will use the section formula: \( P(x,y) = \left(\frac{kx_2+1x_1}{k+1}, \frac{ky_2+1y_1}{k+1}\right) \).
Using the x-coordinates:
\( 1 = \frac{k \times 2 + 1 \times (-2)}{k+1} \)
\( 1 = \frac{2k-2}{k+1} \)
\( k+1 = 2k-2 \)
\( 1+2 = 2k-k \)
\( 3 = k \)
Using the y-coordinates:
\( \frac{-7}{2} = \frac{k \times \left(\frac{-10}{3}\right) + 1 \times (-4)}{k+1} \)
\( \frac{-7}{2} = \frac{\frac{-10k}{3}-4}{k+1} \)
\( -7(k+1) = 2\left(\frac{-10k}{3}-4\right) \)
\( -7k-7 = \frac{-20k}{3}-8 \)
Multiply by 3 to clear the fraction:
\( 3(-7k-7) = -20k-24 \)
\( -21k-21 = -20k-24 \)
\( -21+24 = -20k+21k \)
\( 3 = k \)
Since both x and y coordinates give the same value of k=3, the ratio is k:1, which is 3:1. This means the point divides the segment internally.
The line segment AB with point P is shown below:

In simple words: To find the ratio in which a point divides a line, we assume the ratio is k:1. Then, we use the section formula with the given point and the endpoints. We solve for k using both the x and y coordinates. If k is positive, it's an internal division; if negative, it's external.

🎯 Exam Tip: Always solve for 'k' using both the x and y coordinates. If they yield different values, it means the three points are not collinear. If the values match, that's your ratio. Be careful with fractional coordinates.

Question 16. In what ratio is the line joining the points
(i) (2, -3) and (5, 6) divided by the x-axis ;
(ii) (3, -6) and (-6, 8) divided by the y-axis?
Answer: When a line segment is divided by an axis, the dividing point lies on that axis.
(i) Divided by the x-axis:
Let the line segment joining A(2, -3) and B(5, 6) be divided by a point P(x, 0) on the x-axis in the ratio k:1. (A point on the x-axis always has its y-coordinate as 0).
Using the section formula for the y-coordinate:
\( y = \frac{ky_2+1y_1}{k+1} \)
\( 0 = \frac{k \times 6 + 1 \times (-3)}{k+1} \)
\( 0 = \frac{6k-3}{k+1} \)
\( 0 = 6k-3 \)
\( 3 = 6k \)
\( k = \frac{3}{6} = \frac{1}{2} \)
The ratio is k:1, which is \( \frac{1}{2}:1 \), or 1:2. This is an internal division because k is positive. The x-axis cuts the line segment into two parts with this ratio.
The line segment AB divided by the x-axis at P is shown below:

(ii) Divided by the y-axis:
Let the line segment joining A(3, -6) and B(-6, 8) be divided by a point P(0, y) on the y-axis in the ratio k:1. (A point on the y-axis always has its x-coordinate as 0).
Using the section formula for the x-coordinate:
\( x = \frac{kx_2+1x_1}{k+1} \)
\( 0 = \frac{k \times (-6) + 1 \times 3}{k+1} \)
\( 0 = \frac{-6k+3}{k+1} \)
\( 0 = -6k+3 \)
\( 6k = 3 \)
\( k = \frac{3}{6} = \frac{1}{2} \)
The ratio is k:1, which is \( \frac{1}{2}:1 \), or 1:2. This is also an internal division.
In simple words: When a line crosses the x-axis, its y-coordinate is zero. When it crosses the y-axis, its x-coordinate is zero. We use the section formula with these zero coordinates to find the ratio in which the axis divides the line.

🎯 Exam Tip: Remember that a point on the x-axis has coordinates (x, 0) and a point on the y-axis has coordinates (0, y). Use the section formula with the coordinate that is zero to quickly solve for the ratio 'k'.

Question 17. Find the ratio in which the axes divide the line joining the points (2, 5) and (1, 9).
Answer: We need to find two ratios: one for division by the x-axis and another for division by the y-axis. Let the given points be A(2, 5) and B(1, 9). We assume a point P divides AB in the ratio k:1.
1. Division by the x-axis:
A point on the x-axis has coordinates (x, 0).
Using the section formula for the y-coordinate:
\( 0 = \frac{k \times 9 + 1 \times 5}{k+1} \)
\( 0 = \frac{9k+5}{k+1} \)
\( 0 = 9k+5 \)
\( 9k = -5 \)
\( k = -\frac{5}{9} \)
Since k is negative, the division is external. The ratio is \( \frac{5}{9}:1 \), or 5:9 externally. This means the x-axis intersects the line segment extended outside the original segment.
The line segment AB with external division by the x-axis is shown below:

2. Division by the y-axis:
A point on the y-axis has coordinates (0, y).
Using the section formula for the x-coordinate:
\( 0 = \frac{k \times 1 + 1 \times 2}{k+1} \)
\( 0 = \frac{k+2}{k+1} \)
\( 0 = k+2 \)
\( k = -2 \)
Since k is negative, the division is external. The ratio is \( 2:1 \) externally.
In simple words: When the x-axis cuts a line, the y-coordinate of that point is zero. When the y-axis cuts a line, the x-coordinate is zero. We use these facts with the section formula to find the ratios, and a negative ratio means the cut happens outside the line segment.

🎯 Exam Tip: Remember to solve for division by both axes separately. A negative value of 'k' indicates external division, which means the dividing point lies outside the line segment. Always state whether the division is internal or external.

Question 18. Show by using section formula that the point (3, -2 ),(5, 2) and (8, 8) are collinear.
Answer: Three points are collinear if one point divides the line segment formed by the other two points in some ratio (k:1). If we can find a consistent ratio 'k' for both the x and y coordinates, then the points are collinear.
Let the given points be A(3, -2), B(5, 2), and C(8, 8).
Assume that point B divides the line segment AC in the ratio k:1.
Using the section formula for point B(5, 2) and endpoints A(3, -2), C(8, 8):
For x-coordinate:
\( 5 = \frac{k \times 8 + 1 \times 3}{k+1} \)
\( 5(k+1) = 8k+3 \)
\( 5k+5 = 8k+3 \)
\( 5-3 = 8k-5k \)
\( 2 = 3k \)
\( k = \frac{2}{3} \)
For y-coordinate:
\( 2 = \frac{k \times 8 + 1 \times (-2)}{k+1} \)
\( 2 = \frac{8k-2}{k+1} \)
\( 2(k+1) = 8k-2 \)
\( 2k+2 = 8k-2 \)
\( 2+2 = 8k-2k \)
\( 4 = 6k \)
\( k = \frac{4}{6} = \frac{2}{3} \)
Since both the x-coordinate and y-coordinate yield the same value of k \( = \frac{2}{3} \), it means that B divides AC in the ratio 2:3. Therefore, the points A, B, and C lie on the same line, proving they are collinear. This consistent ratio is key for collinearity.
The collinear points A, B, C are shown below:

In simple words: Three points are in a straight line if one point can be found by dividing the line segment between the other two points in a specific ratio. If the ratio calculated using the x-coordinates is the same as the ratio calculated using the y-coordinates, then the points are on the same line.

🎯 Exam Tip: To prove collinearity using the section formula, assume one point divides the segment formed by the other two in a k:1 ratio. If both x and y coordinates yield the same 'k', the points are collinear. If 'k' is different, they are not.

Question 19. Find the centroid of the triangle whose angular points are (-4, 6),(2, -2) and (2, 5) respectively.
Answer: The centroid of a triangle is the point where the three medians of the triangle intersect. The coordinates of the centroid G for a triangle with vertices \( (x_1, y_1), (x_2, y_2), \) and \( (x_3, y_3) \) are given by the formula: \( G = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right) \). This formula effectively finds the average position of the vertices.
Given vertices are A(-4, 6), B(2, -2), and C(2, 5).
Let \( x_1=-4, y_1=6 \)
Let \( x_2=2, y_2=-2 \)
Let \( x_3=2, y_3=5 \)
Coordinates of Centroid G:
\( G = \left(\frac{-4+2+2}{3}, \frac{6+(-2)+5}{3}\right) \)
\( = \left(\frac{0}{3}, \frac{6-2+5}{3}\right) \)
\( = \left(0, \frac{9}{3}\right) \)
\( = (0, 3) \)
The centroid of the triangle is (0, 3).
The triangle with its centroid is shown below:

In simple words: The centroid is like the triangle's balancing point. You find it by adding all the x-coordinates together and dividing by three, then do the same for all the y-coordinates. This gives you one single point.

🎯 Exam Tip: The centroid formula is a direct average of the coordinates. Ensure you add all three x-values and y-values correctly, being careful with negative signs, before dividing by three.

Question 20. If \( (x_1, y_1) = (2, 3); x_2 = 3 \) and \( y_3 = -2 \) and G is (0, 0), find \( y_2 \) and \( x_3 \).
Answer: We are given some coordinates of the vertices of a triangle and its centroid G(0, 0). We need to find the missing coordinates \( y_2 \) and \( x_3 \). We will use the centroid formula: \( G(x,y) = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right) \).
Given:
\( (x_1, y_1) = (2, 3) \)
\( x_2 = 3 \)
\( y_3 = -2 \)
Centroid G = (0, 0)
From the x-coordinate of the centroid:
\( 0 = \frac{x_1+x_2+x_3}{3} \)
\( 0 = \frac{2+3+x_3}{3} \)
\( 0 = \frac{5+x_3}{3} \)
\( 0 = 5+x_3 \)
\( x_3 = -5 \)
From the y-coordinate of the centroid:
\( 0 = \frac{y_1+y_2+y_3}{3} \)
\( 0 = \frac{3+y_2+(-2)}{3} \)
\( 0 = \frac{3+y_2-2}{3} \)
\( 0 = \frac{1+y_2}{3} \)
\( 0 = 1+y_2 \)
\( y_2 = -1 \)
Thus, the missing coordinates are \( x_3 = -5 \) and \( y_2 = -1 \). Finding these missing pieces helps complete the picture of the triangle's structure.
In simple words: The centroid is the average of all x-coordinates and all y-coordinates. If we know the centroid and some of the corner points, we can use these averages to work backward and find any missing corner coordinates.

🎯 Exam Tip: Treat the x and y coordinates of the centroid formula as two separate equations. This allows you to solve for each unknown variable independently. Be careful with algebra, especially when isolating the unknown terms.

Question 21. Find the coordinates of the in-centre of the triangle whose vertices are (-36, 7) and (20, 7) and (0, -8).
Answer: Let the vertices of the triangle be A(\(x_1\), \(y_1\)) = (-36, 7), B(\(x_2\), \(y_2\)) = (20, 7), and C(\(x_3\), \(y_3\)) = (0, -8).
First, we calculate the lengths of the sides of the triangle:
Side c (length AB):
\( c = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
\( c = \sqrt{(20 - (-36))^2 + (7 - 7)^2} \)
\( c = \sqrt{(56)^2 + (0)^2} \)
\( c = \sqrt{3136} = 56 \)
Side b (length AC):
\( b = \sqrt{(x_3 - x_1)^2 + (y_3 - y_1)^2} \)
\( b = \sqrt{(0 - (-36))^2 + (-8 - 7)^2} \)
\( b = \sqrt{(36)^2 + (-15)^2} \)
\( b = \sqrt{1296 + 225} = \sqrt{1521} = 39 \)
Side a (length BC):
\( a = \sqrt{(x_3 - x_2)^2 + (y_3 - y_2)^2} \)
\( a = \sqrt{(0 - 20)^2 + (-8 - 7)^2} \)
\( a = \sqrt{(-20)^2 + (-15)^2} \)
\( a = \sqrt{400 + 225} = \sqrt{625} = 25 \)
The formula for the in-centre I(\(X, Y\)) of a triangle with vertices A(\(x_1, y_1\)), B(\(x_2, y_2\)), C(\(x_3, y_3\)) and corresponding side lengths a, b, c is:
\( I = \left(\frac{ax_1 + bx_2 + cx_3}{a+b+c}, \frac{ay_1 + by_2 + cy_3}{a+b+c}\right) \)
Now, we substitute the values:
\( X = \frac{25 \times (-36) + 39 \times 20 + 56 \times 0}{25 + 39 + 56} \)
\( X = \frac{-900 + 780 + 0}{120} \)
\( X = \frac{-120}{120} = -1 \)
\( Y = \frac{25 \times 7 + 39 \times 7 + 56 \times (-8)}{25 + 39 + 56} \)
\( Y = \frac{175 + 273 - 448}{120} \)
\( Y = \frac{448 - 448}{120} = \frac{0}{120} = 0 \)
Thus, the coordinates of the in-centre are I(-1, 0). The in-centre is the point where the angle bisectors of the triangle meet, and it is always equidistant from all three sides.
In simple words: First, we find the length of each side of the triangle using the distance formula. Then, we use a special formula for the in-centre that involves these side lengths and the coordinates of the triangle's corners. After putting all the numbers into the formula and doing the calculations, we get the exact spot for the in-centre.

🎯 Exam Tip: Remember to correctly identify the opposite side length for each vertex (\(a\) for \(A\), \(b\) for \(B\), \(c\) for \(C\)) and be careful with negative signs during calculations. A diagram can sometimes help visualize the triangle, even if not strictly required for the calculation.