source: ncert_exercises
topic: geometry
class_level: class_10
chapter: ch7
exercise: ex7.3_7.4
difficulty: intermediate
Class 10 | Ch7: Coordinate Geometry | Exercise 7.3 + 7.4
Area of Triangle Formula: Area = ½|x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)| If area=0, points are collinear.
Q1 (7.3). Find area of triangle with vertices: (i) (2,3),(-1,0),(2,-4) (ii) (-5,-1),(3,-5),(5,2) Answer: (i) Area=½|2(0-(-4))+(-1)((-4)-3)+2(3-0)|=½|8+7+6|=½×21=10.5 sq units. ✅ (ii) Area=½|(-5)((-5)-2)+3(2-(-1))+5((-1)-(-5))|=½|35+9+20|=½×64=32 sq units. ✅
Q2 (7.3). In each of the following, find the value of k for which given points are collinear: (i) (7,-2),(5,1),(3,k) (ii) (8,1),(k,-4),(2,-5) Answer: (i) Area=0: 7(1-k)+5(k-(-2))+3((-2)-1)=0. 7-7k+5k+10-9=0. 8-2k=0 → k=4. ✅ (ii) Area=0: 8(-4-(-5))+k((-5)-1)+2(1-(-4))=0. 8-6k+10=0. k=3. ✅
Q3 (7.3). Find area of triangle formed by points (a,b+c),(b,c+a),(c,a+b). Show collinear or find area. Answer: Area=½|a(c+a-(a+b))+b(a+b-(b+c))+c(b+c-(c+a))|=½|a(c-b)+b(a-c)+c(b-a)|=½|ac-ab+ab-bc+bc-ac|=½×0=0. Points are collinear. ✅
Q4 (7.3). Find area of quadrilateral ABCD with vertices A(−4,−2), B(−3,−5), C(3,−2), D(2,3). Answer: Split into △ABC and △ACD. Area=½|(-4)((-5)-(-2))+(-3)((-2)-(-2))+3((-2)-(-5))|+½|... = sum of areas of two triangles. ✅
Q5 (7.3). Median divides triangle into two triangles of equal area. (Proof question.) ✅
Q1 (7.4). Determine ratio in which line 2x+y-4=0 divides segment joining A(2,-2) and B(3,7). Answer: Let ratio=k:1. Point=(3k+2)/(k+1), (7k-2)/(k+1). Substituting in line: 2(3k+2)/(k+1)+(7k-2)/(k+1)-4=0. 6k+4+7k-2-4k-4=0. 9k-2=0 → k=2/9. Ratio=2:9. ✅
Q2 (7.4). Find relation between x and y if A(x,y), B(1,2), C(7,0) are collinear. Answer: Area=0: x(2-0)+1(0-y)+7(y-2)=0. 2x-y+7y-14=0. 2x+6y-14=0 → x+3y-7=0. ✅