Ch 7: Coordinate Geometry (Part 1)
Based on the latest CBSE rationalized syllabus. Master the Distance Formula and Section Formula.
Exercise 7.1 Solutions
💡 Formula: Distance d = √[(x₂ - x₁)² + (y₂ - y₁)²]
Q1. Find the distance between the following pairs of points: (i) (2, 3), (4, 1) (ii) (-5, 7), (-1, 3)
Solution for (i):
Here x₁ = 2, y₁ = 3 and x₂ = 4, y₂ = 1.
d = √[(4 - 2)² + (1 - 3)²]
d = √[(2)² + (-2)²] = √(4 + 4) = √8 = 2√2 units.
Solution for (ii):
x₁ = -5, y₁ = 7 and x₂ = -1, y₂ = 3.
d = √[(-1 - (-5))² + (3 - 7)²] = √[(4)² + (-4)²] = √(16 + 16) = √32 = 4√2 units.
Q3. Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.
Solution:
Let A=(1, 5), B=(2, 3), C=(-2, -11). For collinearity, sum of any two distances must equal the third.
AB = √[(2-1)² + (3-5)²] = √(1² + (-2)²) = √5
BC = √[(-2-2)² + (-11-3)²] = √[(-4)² + (-14)²] = √(16 + 196) = √212
AC = √[(-2-1)² + (-11-5)²] = √[(-3)² + (-16)²] = √(9 + 256) = √265
Since AB + BC ≠ AC, the points are not collinear.
Q7. Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).
Solution:
Let the point on the x-axis be P(x, 0). Given A(2, -5) and B(-2, 9).
Since P is equidistant from A and B, PA = PB.
Therefore, PA² = PB².
(x - 2)² + (0 - (-5))² = (x - (-2))² + (0 - 9)²
(x - 2)² + 25 = (x + 2)² + 81
x² - 4x + 4 + 25 = x² + 4x + 4 + 81
Canceling x² and 4 from both sides:
-4x + 25 = 4x + 81
-8x = 56 ⇒ x = -7
The required point is (-7, 0).
Q8. Find the values of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units.
Solution:
Given PQ = 10. Therefore, PQ² = 100.
(10 - 2)² + (y - (-3))² = 100
8² + (y + 3)² = 100
64 + y² + 6y + 9 = 100
y² + 6y + 73 - 100 = 0
y² + 6y - 27 = 0
Factorizing: y² + 9y - 3y - 27 = 0
y(y + 9) - 3(y + 9) = 0
(y - 3)(y + 9) = 0
Therefore, y = 3 or y = -9.
Exercise 7.2 (Section Formula)
💡 Formula: x = (m₁x₂ + m₂x₁) / (m₁ + m₂), y = (m₁y₂ + m₂y₁) / (m₁ + m₂)
💡 Trick for finding ratio: Always assume the ratio is k : 1.
Q1. Find the coordinates of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2:3.
Solution:
Let P(x, y) be the required point. m₁ = 2, m₂ = 3.
x₁ = -1, y₁ = 7 and x₂ = 4, y₂ = -3.
x = [2(4) + 3(-1)] / (2 + 3) = (8 - 3) / 5 = 5/5 = 1
y = [2(-3) + 3(7)] / (2 + 3) = (-6 + 21) / 5 = 15/5 = 3
The required point is (1, 3).
Q4. Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).
Solution:
Let the required ratio be k : 1. The dividing point is P(-1, 6).
Using the Section Formula for x-coordinate:
-1 = [k(6) + 1(-3)] / (k + 1)
-1(k + 1) = 6k - 3
-k - 1 = 6k - 3
-k - 6k = -3 + 1
-7k = -2 ⇒ k = 2/7
The required ratio is 2 : 7.
Q5. 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.
Solution:
Any point on the x-axis has the y-coordinate as 0. Let the point be P(x, 0) and the ratio be k : 1.
Using the y-coordinate of the Section Formula:
0 = [k(5) + 1(-5)] / (k + 1)
0 = 5k - 5
5k = 5 ⇒ k = 1
So, the line is divided in the ratio 1 : 1 (which means it's the midpoint).
Now, finding x-coordinate:
x = [1(-4) + 1(1)] / (1 + 1) = -3 / 2
The coordinates of the point of division are (-3/2, 0).
Important PYQs
Q. If the points A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a parallelogram, taken in order, find the value of p.
Solution (Midpoint Trick):
We know that the diagonals of a parallelogram bisect each other. Therefore, the midpoint of diagonal AC is the same as the midpoint of diagonal BD.
Midpoint of AC = [ (6+9)/2 , (1+4)/2 ] = (15/2, 5/2)
Midpoint of BD = [ (8+p)/2 , (2+3)/2 ] = ( (8+p)/2, 5/2 )
Equating the x-coordinates:
15/2 = (8 + p) / 2
15 = 8 + p
p = 7