The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices Let ABCD be a square and A (–1, 2) and C (3, 2) be the given vertices. Let the coordinate of vertex B be ( AB = BC ( ABCD is a square) ⇒ AB^2 = BC^2 ⇒ [x – (–1)]^2 + (y – 2)^2 = (x – 3)^2 + (y – 2)^2 (Distance formula) ⇒ (x + 1)^2 = (x – 3)^2 ⇒ x^2 + 2x + 1 = x^2 – 6x + 9 ⇒ 2x + 6x = 9 – 1 ⇒ 8x = 8 ⇒ x = 1 In ΔABC, we have AB ^2 + BC^2 = AC^2 (Pythagoras theorem) ⇒ 2AB^2 = AC^2 ( AB = BC) ⇒ 2[(x – (–1))^2 + (y – 2)^2] = (3 – (–1))^2 + (2 – 2)^2 ⇒ 2[(x + 1)^2 + (y – 2)^2] = (4)^2 + (0)^2 ⇒ 2[(1 + 1)^2 + (y – 2)^2] = 16 ( x = 1) ⇒ 2[ 4 + (y – 2)^2] = 16 ⇒ 8 + 2 (y – 2)^2 = 16 ⇒ 2 (y – 2)^2 = 16 – 8 = 8 ⇒ (y – 2)^2 = 4 ⇒ y – 2 = ± 2 ⇒ y – 2 = 2 or y – 2 = –2 ⇒ y = 4 or y = 0 Thus, the other two vertices of the square ABCD are (1, 4) and (1, 0).
CBSE: 10th Class Sample Test Paper - Coordinate Geometry
Q.1. Find the coordinates of the mid point of the line segment joining the points (4, 3) and (2, 1). Q.2. Find the coordinates of the point which divides the line segment joining the points (1, 3) and (2, 7) in the ratio 3: 4. Q.3. Show that the points (1, 1), (3, -2) and (-1, 4) are collinear. Q.4. Find the centroid of the triangle whose vertices are (3, -5); (- 7, 4) and (10, - 2). Q.5. Find the area of a triangle whose vertices are A (1, 2); B (3, 5) and C (- 4, - 7) Q.6. If the distance of the point P(x, y) from the points A (5, 1) and B (- 1, 5) is equal, show that 3x = 2y. Q.7. In what ratio does the point P (- 4, 6) divide the line segment joining the points A (- 6, 10) and B (3, - 8). Q.8. For what value of m, the points (4, 3), (m, 1) and (1, 9) are collinear. Q.9. Prove that the coordinates of the centroid of a triangle ABC with vertices A(x1, y1), B(x2, y2) and C(x3, y3) are given by (x1+x2+x3)/3 , (y1+y2+y3)/3 Q.10. Prove that the diagonals of a rectangle bisect each other and are of equal length.