Coordinate Geometry
EXERCISE 7.1
CBSE_NCERT_TEXTBOOK
Grade 10
Question:
Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (i) (– 1, – 2), (1, 0), (– 1, 2), (– 3, 0) (ii) (–3, 5), (3, 1), (0, 3), (–1, – 4) (iii) (4, 5), (7, 6), (4, 3), (1, 2)
Step-by-Step Solution
Key Concept: Use the distance formula \(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\) to find side lengths and the slope formula \(m=\frac{y_2-y_1}{x_2-x_1}\) to test parallelism (equal slopes) and perpendicularity (product of slopes = –1). A quadrilateral is a square if all sides are equal and adjacent sides are perpendicular; a rectangle if opposite sides are parallel and adjacent sides are perpendicular; a rhombus if all sides are equal; a parallelogram if opposite sides are parallel (and equal). If any three points are collinear, a quadrilateral cannot be formed.
### Part (i) : Points A(–1, –2), B(1, 0), C(–1, 2), D(–3, 0)
1. Side lengths
- \(AB = \sqrt{(1+1)^2+(0+2)^2}=\sqrt{4+4}=2\sqrt2\)
- \(BC = \sqrt{(-1-1)^2+(2-0)^2}=2\sqrt2\)
- \(CD = \sqrt{(-3+1)^2+(0-2)^2}=2\sqrt2\)
- \(DA = \sqrt{(-1+3)^2+(-2-0)^2}=2\sqrt2\)
All four sides are equal.
2. Slopes of sides
- \(m_{AB}=\frac{0-(-2)}{1-(-1)}=1\)
- \(m_{BC}=\frac{2-0}{-1-1}=-1\)
- \(m_{CD}=\frac{0-2}{-3-(-1)}=1\)
- \(m_{DA}=\frac{-2-0}{-1-(-3)}=-1\)
Opposite sides have equal slopes (AB // CD and BC // DA).
3. Perpendicularity
- \(m_{AB}\times m_{BC}=1\times(-1)=-1\) → adjacent sides are perpendicular.
4. Diagonals
- \(AC = \sqrt{(-1+1)^2+(2+2)^2}=4\)
- \(BD = \sqrt{(1+3)^2+(0-0)^2}=4\)
Diagonals are equal and perpendicular (slopes undefined and 0).
Conclusion: All sides equal, opposite sides parallel, adjacent sides perpendicular ⇒ the quadrilateral is a square.
### Part (ii) : Points A(–3, 5), B(3, 1), C(0, 3), D(–1, –4)
1. Check collinearity of A, B, C
- Slope \(AB = \frac{1-5}{3-(-3)} = -\frac{4}{6}= -\frac{2}{3}\)
- Slope \(BC = \frac{3-1}{0-3}= \frac{2}{-3}= -\frac{2}{3}\)
Since \(m_{AB}=m_{BC}\), points A, B, C lie on the same straight line.
2. Because three of the given points are collinear, they cannot form a closed four‑sided figure.
Conclusion: No quadrilateral is formed (the points are collinear).
### Part (iii) : Points A(4, 5), B(7, 6), C(4, 3), D(1, 2)
1. Slopes
- \(m_{AB}=\frac{6-5}{7-4}=\frac{1}{3}\)
- \(m_{BC}=\frac{3-6}{4-7}=\frac{-3}{-3}=1\)
- \(m_{CD}=\frac{2-3}{1-4}=\frac{-1}{-3}=\frac{1}{3}\)
- \(m_{DA}=\frac{5-2}{4-1}=\frac{3}{3}=1\)
Hence AB // CD (both slope \(\frac13\)) and BC // DA (both slope \(1\)).
2. Side lengths
- \(AB = \sqrt{(7-4)^2+(6-5)^2}=\sqrt{9+1}=\sqrt{10}\)
- \(BC = \sqrt{(4-7)^2+(3-6)^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt2\)
- \(CD = \sqrt{(1-4)^2+(2-3)^2}=\sqrt{9+1}=\sqrt{10}\)
- \(DA = \sqrt{(4-1)^2+(5-2)^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt2\)
Opposite sides are equal.
3. Angles
- Product of adjacent slopes \(m_{AB}\times m_{BC}=\frac13\times1=\frac13
eq-1\); therefore adjacent sides are not perpendicular.
Conclusion: Opposite sides are parallel and equal, but adjacent sides are not perpendicular ⇒ the quadrilateral is a parallelogram (not a rectangle or rhombus).
Overall Answers
- (i) Square
- (ii) No quadrilateral (three points are collinear)
- (iii) Parallelogram
Correct Answer: (i) Square, (ii) No quadrilateral (three points are collinear), (iii) Parallelogram