Chapter 7: Coordinate Geometry
Key Formulas:
• Distance: d = √[(x₂-x₁)² + (y₂-y₁)²]
• Section formula: P = ((mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n))
• Midpoint: M = ((x₁+x₂)/2, (y₁+y₂)/2)
• Area of triangle = ½|x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)|
Exercise 7.1
Q1. Find distance: (i) (2,3) and (4,1) (ii) (-5,7) and (-1,3) (iii) (a,b) and (-a,-b)
Solution(i) √[(4-2)²+(1-3)²] = √(4+4) = 2√2
(ii) √[(-1+5)²+(3-7)²] = √(16+16) = 4√2
(iii) √[(-a-a)²+(-b-b)²] = √(4a²+4b²) = 2√(a²+b²)
Q2. Find distance of (3,-4) from origin.
Solution√(3²+(-4)²) = √(9+16) = √25 = 5 units
Q3. Find point on x-axis equidistant from (5,9) and (-4,6).
SolutionLet point be (x,0)
√[(x-5)²+81] = √[(x+4)²+36]
(x-5)²+81 = (x+4)²+36
x²-10x+106 = x²+8x+52
-18x = -54 → x=3
∴ Point = (3, 0)
Q5. Find x if distance between (x,-1) and (1,3) is 5 units.
Solution√[(1-x)²+(3+1)²] = 5
(1-x)²+16 = 25
(1-x)² = 9
1-x = ±3
x = -2 or x = 4
∴ x = -2 or x = 4
Q7. Find coordinates dividing (-1,7) and (4,-3) in ratio 2:3.
SolutionP = ((2×4+3×(-1))/(2+3), (2×(-3)+3×7)/5)
= ((8-3)/5, (-6+21)/5)
= (5/5, 15/5) = (1, 3)
Q9. Find points of trisection of line joining (4,-1) and (-2,-3).
SolutionFirst point (1:2): ((1×(-2)+2×4)/3, (1×(-3)+2×(-1))/3)
= (6/3, -5/3) = (2, -5/3)
Second point (2:1): ((2×(-2)+1×4)/3, (2×(-3)+1×(-1))/3)
= (0/3, -7/3) = (0, -7/3)
Exercise 7.2
Q1. Find coordinates dividing (-1,7) and (4,-3) internally in ratio 2:3.
SolutionP = ((2×4+3×(-1))/5, (2×(-3)+3×7)/5)
= (5/5, 15/5) = (1, 3)
Q5. Find centroid of ΔABC: A(-1,3), B(1,-1), C(5,1).
SolutionG = ((-1+1+5)/3, (3-1+1)/3)
= (5/3, 3/3) = (5/3, 1)
Exercise 7.3
Q1. Find area: (i) A(7,2), B(9,10), C(3,6) (ii) A(-2,-3), B(3,2), C(-1,-8)
Solution(i) Area = ½|7(10-6)+9(6-2)+3(2-10)|
= ½|28+36-24| = ½|40| = 20 sq. units
(ii) Area = ½|-2(2+8)+3(-8+3)+(-1)(-3-2)|
= ½|-20-15+5| = ½|-30| = 15 sq. units
Q4. Find k if A(-2,3), B(k,1), C(-1,4) are collinear.
SolutionArea = 0 for collinear points
½|-2(1-4)+k(4-3)+(-1)(3-1)| = 0
½|6+k-2| = 0
k+4 = 0
k = -4