Maharashtra Board Class 10 Maths Chapter 5 Co ordinate Geometry Set 5.2 Solutions

Question 1. Find the co-ordinates of point P if P divides the line segment joining the points A (-1, 7) and B (4, -3) in the ratio 2:3.
Answer:
Solution:
Let the co-ordinates of point P be (x, y) and A (x\(_{1}\), y\(_{1}\)) B (x\(_{2}\), y\(_{2}\)) be the given points.
Here, x\(_{1}\) = -1, y\(_{1}\) = 7, x\(_{2}\) = 4, y\(_{2}\) = -3, m = 2, n = 3
.. By section formula,
x = \(\frac{mx_{2}+nx_{1}}{m+n}\) = \(\frac{2(4)+3(-1)}{2+3}\)
= \(\frac{8-3}{5}\) = \(\frac{5}{5}\) = 1
y = \(\frac{my_{2}+ny_{1}}{m+n}\) = \(\frac{2(-3)+3(7)}{2+3}\)
= \(\frac{-6+21}{5}\) = \(\frac{15}{5}\) = 3
The co-ordinates of point P are (1, 3).
.. The co-ordinates of point P are (1,3).
In simple words: To find the coordinates of point P, we use the section formula with the given coordinates of A and B, and the ratio m:n. We calculate x and y coordinates separately using the formula to get the final point P.

🎯 Exam Tip: Remember the section formula \( (x, y) = \left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right) \) and accurately substitute the values to avoid calculation errors.

Question 2. In each of the following examples find the co-ordinates of point A which divides segment PQ in the ratio a : b.
(i) P (-3, 7), Q (1, -4), a : b = 2:1
(ii) P (-2, -5), Q (4, 3), a : b = 3:4
(iii) P (2, 6), Q (-4, 1), a : b = 1 : 2
Answer:
Solution:
Let the co-ordinates of point A be (x, y).
(i) Let P (x\(_{1}\), y\(_{1}\)), Q (x\(_{2}\), y\(_{2}\)) be the given points.
Here, x\(_{1}\) = -3, y\(_{1}\) = 7, x\(_{2}\) = 1, y\(_{2}\) = -4, a = 2, b = 1
.. By section formula,
x = \(\frac{ax_{2}+bx_{1}}{a+b}\) = \(\frac{2(1)+1(-3)}{2+1}\)
= \(\frac{2-3}{3}\) = \(\frac{-1}{3}\)
y = \(\frac{ay_{2}+by_{1}}{a+b}\) = \(\frac{2(-4)+1(7)}{2+1}\)
= \(\frac{-8+7}{3}\) = \(\frac{-1}{3}\)
.. The co-ordinates of point A are \((\frac{-1}{3},\frac{-1}{3})\).
(ii) Let P (x\(_{1}\),y\(_{1}\)), Q (x\(_{2}\), y\(_{2}\)) be the given points.
Here, x\(_{1}\) = -2, y\(_{1}\) = -5, x\(_{2}\) = 4, y\(_{2}\) = 3, a = 3, b = 4
By section formula,
x = \(\frac{ax_{2}+bx_{1}}{a+b}\) = \(\frac{3(4)+4(-2)}{3+4}\)
= \(\frac{12-8}{7}\) = \(\frac{4}{7}\)
y = \(\frac{ay_{2}+by_{1}}{a+b}\) = \(\frac{3(3)+4(-5)}{3+4}\)
= \(\frac{9-20}{7}\) = \(\frac{-11}{7}\)
.. The co-ordinates of point A are \((\frac{4}{7},\frac{-11}{7})\).
(iii) Let P (x\(_{1}\), y\(_{1}\)), Q (x\(_{2}\), y\(_{2}\)) be the given points.
Here,x\(_{1}\) = 2,y\(_{1}\) = 6, x\(_{2}\) = -4, y\(_{2}\) = 1, a = 1,b = 2
.. By section formula,
x = \(\frac{ax_{2}+bx_{1}}{a+b}\) = \(\frac{1(-4)+2(2)}{1+2}\)
= \(\frac{-4+4}{3}\)
= 0
y = \(\frac{ay_{2}+by_{1}}{a+b}\) = \(\frac{1(1)+2(6)}{1+2}\)
= \(\frac{1+12}{3}\)
= \(\frac{13}{3}\)
.. The co-ordinates of point A are \((0, \frac{13}{3})\).
In simple words: For each sub-part, we apply the section formula using the given coordinates of points P and Q, and the ratio a:b, to find the coordinates (x, y) of point A. This involves calculating both the x and y coordinates separately.

🎯 Exam Tip: Pay close attention to the signs of the coordinates and the ratio values when substituting them into the section formula to avoid errors in calculation.

Question 3. Find the ratio in which point T (-1, 6) divides the line segment joining the points P (-3,10) and Q (6, -8).
Answer:
Solution:
Let P (x\(_{1}\), y\(_{1}\)), Q (x\(_{2}\), y\(_{2}\)) and T (x, y) be the given points.
Here, x\(_{1}\) = -3, y\(_{1}\) = 10, x\(_{2}\) = 6, y\(_{2}\) = -8, x = -1, y = 6
.. By section formula,
x = \(\frac{mx_{2}+nx_{1}}{m+n}\)
-1 = \(\frac{m(6)+n(-3)}{m+n}\)

\(\implies\) -1(m+n) = 6m - 3n

\(\implies\) -m-n = 6m - 3n

\(\implies\) -n+3n = 6m + m

\(\implies\) 2n = 7m

\(\implies\) \(\frac{m}{n}\) = \(\frac{2}{7}\)
.. m:n=2:7
.. Point T divides seg PQ in the ratio 2 : 7.
In simple words: We use the section formula by setting the given coordinates of point T equal to the formula for x and y. This allows us to form an equation that helps us solve for the ratio m:n in which T divides the segment PQ.

🎯 Exam Tip: When finding the ratio, you can use either the x-coordinate or the y-coordinate with the section formula; both should yield the same ratio. Choose the one that simplifies calculations.

Question 4. Point P is the centre of the circle and AB is a diameter. Find the co-ordinates of point B if co-ordinates of point A and P are (2, -3) and (-2,0) respectively.
Answer:
Solution:
Let A (x\(_{1}\), y\(_{1}\)), B (x\(_{2}\), y\(_{2}\)) and P (x, y) be the given points.
Here, x\(_{1}\) = 2, y\(_{1}\) =-3,
x = -2, y = 0

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक वृत्त को दर्शाता है जिसमें AB एक व्यास है। A के निर्देशांक (2,-3) और P (-2,0) हैं, जहां P वृत्त का केंद्र है। P व्यास AB का मध्यबिंदु है।

Point P is the midpoint of seg AB.
.. By midpoint formula,
x = \(\frac{x_{1}+x_{2}}{2}\)
-2 = \(\frac{2+x_{2}}{2}\)

\(\implies\) -4 = 2+x\(_{2}\)

\(\implies\) x\(_{2}\) = -4-2 = -6
y = \(\frac{y_{1}+y_{2}}{2}\)
0 = \(\frac{-3+y_{2}}{2}\)

\(\implies\) -3 + y\(_{2}\) = 0

\(\implies\) y\(_{2}\) = 3
.. The co-ordinates of point B are (-6,3).
In simple words: Since P is the center and AB is the diameter, P is the midpoint of segment AB. We use the midpoint formula with the given coordinates of A and P to find the unknown coordinates of B by solving for x\(_{2}\) and y\(_{2}\).

🎯 Exam Tip: Recognize that the center of a circle is the midpoint of its diameter. This understanding is key to applying the correct midpoint formula for coordinates.

Question 5. Find the ratio in which point P (k, 7) divides the segment joining A (8, 9) and B (1,2). Also find k.
Answer:
Solution:
Let A (x\(_{1}\), y\(_{1}\)), B (x\(_{2}\), y\(_{2}\)) and P (x, y) be the given points.
Here, x\(_{1}\) = 8, y\(_{1}\) = 9, x\(_{2}\) = 1, y\(_{2}\) = 2, x = k, y = 7
.. By section formula,
y = \(\frac{my_{2}+ny_{1}}{m+n}\)
7 = \(\frac{2m+9n}{m+n}\)

\(\implies\) 7(m+n) = 2m + 9n

\(\implies\) 7m+7n = 2m + 9n

\(\implies\) 5m = 2n

\(\implies\) \(\frac{m}{n}\) = \(\frac{2}{5}\)
.. m:n=2:5
x = \(\frac{mx_{2}+nx_{1}}{m+n}\)
k = \(\frac{2(1)+5(8)}{2+5}\)
= \(\frac{2+40}{7}\)
= \(\frac{42}{7}\)
= 6
.. Point P divides seg AB in the ratio 2 : 5, and the value of k is 6.
In simple words: First, use the known y-coordinates in the section formula to find the ratio m:n. Once the ratio is determined, use it along with the x-coordinates in the section formula to solve for the unknown value of k.

🎯 Exam Tip: When both the ratio and an unknown coordinate (like 'k') need to be found, always use the known coordinates first to establish the ratio, then apply the ratio to find the unknown coordinate.

Question 6. Find the co-ordinates of midpoint of the segment joining the points (22, 20) and (0,16).
Answer:
Solution:
Let A (x\(_{1}\), y\(_{1}\)) = A (22, 20),
B (x\(_{2}\),y\(_{2}\)) = B (0, 16)
Let the co-ordinates of the midpoint be P (x,y).
.. By midpoint formula,
x = \(\frac{x_{1}+x_{2}}{2}\) = \(\frac{22+0}{2}\) = 11
y = \(\frac{y_{1}+y_{2}}{2}\) = \(\frac{20+16}{2}\) = \(\frac{36}{2}\) = 18
The co-ordinates of the midpoint of the segment joining (22, 20) and (0, 16) are (11,18).
In simple words: To find the midpoint's coordinates, we sum the x-coordinates and divide by two, then sum the y-coordinates and divide by two. This gives us the average position for both axes.

🎯 Exam Tip: The midpoint formula is a special case of the section formula where the ratio is 1:1. It's simply the average of the respective coordinates.

Question 7. Find the centroids of the triangles whose vertices are given below.
(i) (-7, 6), (2,-2), (8, 5)
(ii) (3, -5), (4, 3), (11,-4)
(iii) (4, 7), (8, 4), (7, 11)
Answer:
Solution:
(i) Let A (x\(_{1}\), y\(_{1}\)) = A (-7, 6),
B (x\(_{2}\), y\(_{2}\)) = B (2, -2),
C (x\(_{3}\), y\(_{3}\)) = C(8, 5)
.. By centroid formula,
x = \(\frac{x_{1}+x_{2}+x_{3}}{3}\)
= \(\frac{-7+2+8}{3}\) = \(\frac{3}{3}\) = 1
y = \(\frac{y_{1}+y_{2}+y_{3}}{3}\)
= \(\frac{6-2+5}{3}\) = \(\frac{9}{3}\) = 3
.. The co-ordinates of the centroid are (1,3).
(ii) Let A (x\(_{1}\) y\(_{1}\)) = A (3, -5),
В (x\(_{2}\), y\(_{2}\)) = B (4, 3),
C(x\(_{3}\), y\(_{3}\)) = C(11,-4)
.. By centroid formula,
x = \(\frac{x_{1}+x_{2}+x_{3}}{3}\)
= \(\frac{3+4+11}{3}\) = \(\frac{18}{3}\) = 6
y = \(\frac{y_{1}+y_{2}+y_{3}}{3}\)
= \(\frac{-5+3-4}{3}\) = \(\frac{-6}{3}\) = -2
.. The co-ordinates of the centroid are (6, -2).
(iii) Let A (x\(_{1}\), y\(_{1}\)) = A (4, 7),
В (x\(_{2}\), y\(_{2}\)) = B (8,4),
C (x\(_{3}\), y\(_{3}\)) = C(7,11)
.. By centroid formula,
x = \(\frac{x_{1}+x_{2}+x_{3}}{3}\)
= \(\frac{4+8+7}{3}\) = \(\frac{19}{3}\)
y = \(\frac{y_{1}+y_{2}+y_{3}}{3}\)
= \(\frac{7+4+11}{3}\) = \(\frac{22}{3}\)
.. The co-ordinates of the centroid are \((\frac{19}{3},\frac{22}{3})\).
In simple words: The centroid of a triangle is found by averaging the x-coordinates of all three vertices and averaging the y-coordinates of all three vertices separately. This gives a single point that is the triangle's center of mass.

🎯 Exam Tip: The centroid formula is a straightforward average of coordinates. Ensure correct addition of positive and negative numbers for accuracy in calculations.

Question 8. In ∆ABC, G (-4, -7) is the centroid. If A (-14, -19) and B (3, 5), then find the co-ordinates of C.
Answer:
Solution:
G (x, y) = G (-4, -7),
A (x\(_{1}\), y\(_{1}\)) = A (-14, -19),
B(x\(_{2}\), y\(_{2}\)) = B(3,5)
Let the co-ordinates of point C be (x\(_{3}\), y\(_{3}\)).
G is the centroid.
By centroid formula,
x = \(\frac{x_{1}+x_{2}+x_{3}}{3}\)
-4 = \(\frac{-14+3+x_{3}}{3}\)

\(\implies\) -12 = -11+x\(_{3}\)

\(\implies\) x\(_{3}\) = -12+11

\(\implies\) x\(_{3}\) = -1
y = \(\frac{y_{1}+y_{2}+y_{3}}{3}\)
-7 = \(\frac{-19+5+y_{3}}{3}\)

\(\implies\) -21 = -14 + y\(_{3}\)

\(\implies\) y\(_{3}\) = -21 +14

\(\implies\) y\(_{3}\) = -7
.. The co-ordinates of point C are (-1, – 7).
In simple words: Given the centroid and two vertices, we can use the centroid formula to find the missing vertex. By setting the centroid's coordinates equal to the average of all three vertices' coordinates, we can solve for the unknown x and y values of the third vertex.

🎯 Exam Tip: When working backward from the centroid to find a missing vertex, carefully apply inverse operations (multiplication then subtraction/addition) to isolate the unknown coordinate.

Question 9. A (h, -6), B (2, 3) and C (-6, k) are the co-ordinates of vertices of a triangle whose centroid is G (1,5). Find h and k.
Answer:
Solution:
A(x\(_{1}\),y\(_{1}\)) = A(h, -6),
В (x\(_{2}\), y\(_{2}\)) = B(2, 3),
C (x\(_{3}\), y\(_{3}\)) = C (-6, k)
.. centroid G (x, y) = G (1, 5)
G is the centroid.
By centroid formula,
x = \(\frac{x_{1}+x_{2}+x_{3}}{3}\)
1 = \(\frac{h+2+(-6)}{3}\)

\(\implies\) 3 = h+2-6

\(\implies\) 3 = h-4

\(\implies\) h = 3 + 4

\(\implies\) h = 7
y = \(\frac{y_{1}+y_{2}+y_{3}}{3}\)
5 = \(\frac{-6+3+k}{3}\)

\(\implies\) 15 = -3 + k

\(\implies\) k = 15 + 3

\(\implies\) k = 18
.. h = 7 and k = 18
In simple words: We use the centroid formula, substituting the given centroid coordinates and the partially known vertex coordinates. This creates two separate equations, one for 'h' (from x-coordinates) and one for 'k' (from y-coordinates), which we then solve.

🎯 Exam Tip: Treat the calculation for 'h' and 'k' as two independent problems derived from the x and y components of the centroid formula. Organize your steps clearly for each variable.

Question 10. Find the co-ordinates of the points of trisection of the line segment AB with A (2,7) and B (-4, -8).
Answer:
Solution:
A (2, 7), B (-4, -8)
Suppose the points P and Q trisect seg AB.
.. AP = PQ = QB

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक रेखा खंड AB को दर्शाता है जिसके बिंदु A(2,7) और B(-4,-8) हैं। बिंदु P और Q इस रेखा खंड को तीन बराबर भागों में विभाजित करते हैं, इस प्रकार AP = PQ = QB होता है।

Let AP = PQ = QB = x
\(\frac{AP}{PB}\) = \(\frac{AP}{PQ + QB}\)

\(\implies\) \(\frac{AP}{PB}\) = \(\frac{x}{x + x}\) = \(\frac{x}{2x}\) = \(\frac{1}{2}\)
.. Point P divides seg AB in the ratio 1:2.
.. By section formula,
x co-ordinate of P = \(\frac{mx_{2}+nx_{1}}{m+n}\)
= \(\frac{1(-4)+2(2)}{1+2}\)
= \(\frac{-4+4}{3}\)
= 0
y co-ordinate of P = \(\frac{my_{2}+ny_{1}}{m+n}\)
= \(\frac{1(-8)+2(7)}{1+2}\)
= \(\frac{-8+14}{3}\) = \(\frac{6}{3}\) = 2
Co-ordinates of P are (0, 2).
Point Q is the midpoint of PB.
By midpoint formula,
x co-ordinate of Q = \(\frac{x_{1}+x_{2}}{2}\)
= \(\frac{0+(-4)}{2}\) = \(\frac{-4}{2}\) = -2
y co-ordinate of Q = \(\frac{y_{1}+y_{2}}{2}\)
= \(\frac{2+(-8)}{2}\)
= \(\frac{2-8}{2}\) = \(\frac{-6}{2}\) = -3
Co-ordinates of Q are (-2, -3).
.. The co-ordinates of the points of trisection seg AB are (0,2) and (-2, -3).
In simple words: To find the points of trisection, we first determine the ratio in which the first point P divides the segment AB (1:2). We use the section formula to find P. Then, we recognize that the second point Q is the midpoint of PB, so we use the midpoint formula to find Q.

🎯 Exam Tip: For trisection, remember that the first point divides the segment in a 1:2 ratio, and the second point divides it in a 2:1 ratio (or is the midpoint of the remaining segment after the first point). Using the midpoint formula for the second point (Q) often simplifies calculations.

Question 11. If A (-14, -10), B (6, -2) are given, find the co-ordinates of the points which divide segment AB into four equal parts.
Answer:
Solution:
Let the points C, D and E divide seg AB in four equal parts.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक रेखा खंड AB को दर्शाता है जिसके बिंदु A(-14,-10) और B(6,-2) हैं। बिंदु C, D और E इस रेखा खंड को चार बराबर भागों में विभाजित करते हैं।

Point D is the midpoint of seg AB.
.. By midpoint formula,
x co-ordinate of D = \(\frac{x_{1}+x_{2}}{2}\)
= \(\frac{-14+6}{2}\)
= \(\frac{-8}{2}\) = -4
y co-ordinate of D = \(\frac{y_{1}+y_{2}}{2}\)
= \(\frac{-10-2}{2}\)
= \(\frac{-12}{2}\) = -6
.. Co-ordinates of D are (-4, -6).
Point C is the midpoint of seg AD.
.. By midpoint formula,
x co-ordinate of C = \(\frac{x_{1}+x_{2}}{2}\)
= \(\frac{-14-4}{2}\)
= \(\frac{-18}{2}\) = -9
y co-ordinate of C = \(\frac{y_{1}+y_{2}}{2}\)
= \(\frac{-10-6}{2}\)
= \(\frac{-16}{2}\) = -8
.. Co-ordinates of C are (-9, -8).
Point E is the midpoint of seg DB.
.. By midpoint formula,
x co-ordinate of E = \(\frac{x_{1}+x_{2}}{2}\)
= \(\frac{-4+6}{2}\)
= \(\frac{2}{2}\) = 1
y co-ordinate of E = \(\frac{y_{1}+y_{2}}{2}\)
= \(\frac{-6-2}{2}\)
= \(\frac{-8}{2}\) = -4
.. Co-ordinates of E are (1,-4).
.. The co-ordinates of the points dividing seg AB in four equal parts are C(-9, -8), D(-4, -6) and E(1, – 4).
In simple words: To divide a segment into four equal parts, first find the midpoint of the entire segment (D). Then, find the midpoint of the first half (AD) to get C, and the midpoint of the second half (DB) to get E. Each step involves applying the midpoint formula.

🎯 Exam Tip: When dividing a segment into N equal parts, finding the middle points iteratively using the midpoint formula (e.g., D is midpoint of AB, C is midpoint of AD, E is midpoint of DB) is often simpler than using the section formula for each point directly with different ratios.

Question 12. If A (20, 10), B (0, 20) are given, find the co-ordinates of the points which divide segment AB into five congruent parts.
Answer:
Solution:
Suppose the points C, D, E and F divide seg AB in five congruent parts.
.. AC = CD = DE = EF = FB

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक रेखा खंड AB को दर्शाता है जिसके बिंदु A(20,10) और B(0,20) हैं। बिंदु C, D, E और F इस रेखा खंड को पांच बराबर भागों में विभाजित करते हैं।

\(\frac{AC}{CB}\) = \(\frac{AC}{CD+DE+ EF + FB}\)

\(\implies\) \(\frac{AC}{CB}\) = \(\frac{x}{x+x+x+x}\) = \(\frac{x}{4x}\) = \(\frac{1}{4}\)
Point C divides seg AB in the ratio 1: 4.
By section formula,
x co-ordinate of C = \(\frac{mx_{2}+nx_{1}}{m+n}\)
= \(\frac{1(0)+4(20)}{1+4}\)
= \(\frac{80}{5}\) = 16
y co-ordinate of C = \(\frac{my_{2}+ny_{1}}{m+n}\)
= \(\frac{1(20)+4(10)}{1+4}\)
= \(\frac{20+40}{5}\)
= \(\frac{60}{5}\) = 12
.. co-ordinates of C are (16, 12).
E is the midpoint of seg CB.
By midpoint formula,
x co-ordinate of E = \(\frac{x_{1}+x_{2}}{2}\)
= \(\frac{16+0}{2}\) = \(\frac{16}{2}\) = 8
y co-ordinate of E = \(\frac{y_{1}+y_{2}}{2}\)
= \(\frac{12+20}{2}\) = \(\frac{32}{2}\) = 16
.. co-ordinates of E are (8, 16).
D is the midpoint of seg CE.
x co-ordinate of D= \(\frac{x_{1}+x_{2}}{2}\)
= \(\frac{16+8}{2}\) = \(\frac{24}{2}\) = 12
y co-ordinate ofD = \(\frac{y_{1}+y_{2}}{2}\)
= \(\frac{12+16}{2}\) = \(\frac{28}{2}\) = 14
.. co-ordinates of D are (12, 14).
F is the midpoint of seg EB.
x co-ordinate of F = \(\frac{x_{1}+x_{2}}{2}\)
= \(\frac{8+0}{2}\) = 4
y co-ordinate of F = \(\frac{y_{1}+y_{2}}{2}\)
= \(\frac{16+20}{2}\) = \(\frac{36}{2}\) = 18
.. co-ordinates of F are (4, 18).
.. The co-ordinates of the points dividing seg AB in five congruent parts are C (16, 12), D (12, 14), E (8, 16) and F (4, 18).
In simple words: To divide a segment into five equal parts, we can use a combination of the section formula and the midpoint formula. First, find point C using the section formula with ratio 1:4. Then, use the midpoint formula iteratively: E is the midpoint of CB, D is the midpoint of CE, and F is the midpoint of EB.

🎯 Exam Tip: For divisions into many equal parts, strategic use of the section formula for the first point (e.g., C divides AB in 1:4) and then repeatedly applying the midpoint formula for subsequent points (e.g., D is midpoint of AC, E is midpoint of CB) can be more efficient than using the section formula for every single point with unique ratios.

Question 1. A (15, 5), B (9, 20) and A-P-B. Find the ratio in which point P (11, 15) divides segment AB. Find the ratio using x and y co-ordinates. Write the conclusion. (Textbook pg. no. 113)
Answer:
Solution:
Suppose point P (11,15) divides segment AB in the ratio m: n.
By section formula,
x = \(\frac{mx_{2}+nx_{1}}{m+n}\)
11 = \(\frac{9m+15n}{m+n}\)

\(\implies\) 11m+11n = 9m + 15n

\(\implies\) 2m = 4n

\(\implies\) \(\frac{m}{n}\) = \(\frac{4}{2}\) = \(\frac{2}{1}\) = 2:1
y = \(\frac{my_{2}+ny_{1}}{m+n}\)
15 = \(\frac{20m+5n}{m+n}\)

\(\implies\) 15m+15n = 20m + 5n

\(\implies\) 5m = 10n

\(\implies\) \(\frac{m}{n}\) = \(\frac{10}{5}\) = \(\frac{2}{1}\) = 2:1
.. Point P divides seg AB in the ratio 2 : 1.
The ratio obtained by using x and y co-ordinates is the same.
In simple words: To find the ratio, we apply the section formula for both x and y coordinates, setting them equal to the given coordinates of P. Solving these equations for m:n will show that both the x and y calculations yield the same ratio, confirming our result.

🎯 Exam Tip: Always verify the ratio using both x and y coordinates if possible; if they don't match, there's likely a calculation error. This serves as a good self-check mechanism.

Question 2. External division: (Textbook pg. no. 115)
Suppose point R divides seg PQ externally in the ratio 3:1.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक रेखा खंड PQ को दर्शाता है, जिसमें बिंदु P के निर्देशांक (x1, y1) और Q के निर्देशांक (x2, y2) हैं। एक बिंदु R इस खंड को बाह्य रूप से इस प्रकार विभाजित करता है कि PR और QR का अनुपात 3:1 है, और R के निर्देशांक (x, y) हैं।

\(\frac{PR}{QR}\) = \(\frac{3}{1}\)
Let the common multiple be k.
Let PR = 3k and QR = k
Now, PR = PQ + QR ... [P - Q - R]
.. 3k = PQ + k
\(\frac{PQ}{QR}\) = \(\frac{2k}{k}\) = \(\frac{2}{1}\)
.. Point Q divides seg PR in the ratio 2 : 1 internally.
Thus, we can find the co-ordinates of point R, when co-ordinates of points P and Q are given.
In simple words: If point R externally divides PQ in a 3:1 ratio, it implies that Q divides PR internally in a 2:1 ratio. This transformation allows us to use the familiar internal section formula to find the coordinates of R, given P and Q.

🎯 Exam Tip: External division problems can often be converted into internal division problems by understanding the relationship between the points and ratios. This simplifies the application of the section formula by working with positive ratios.