Home
Class 12
MATHS
If the vertex of square OABC are O(0,0),...

If the vertex of square OABC are `O(0,0),A(a, b),B(8,-6),C(c,d)` then `|a|+|b|=`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to find the values of \(a\) and \(b\) such that the vertices of the square \(OABC\) are given as \(O(0,0)\), \(A(a,b)\), \(B(8,-6)\), and \(C(c,d)\). We will use the properties of squares and the coordinates of the given points. ### Step-by-Step Solution: 1. **Understanding the Properties of a Square**: A square has all sides of equal length and the diagonals are equal and bisect each other at right angles. 2. **Finding the Length of Side \(OB\)**: The length of side \(OB\) can be calculated using the distance formula: \[ OB = \sqrt{(8 - 0)^2 + (-6 - 0)^2} = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \] 3. **Finding the Coordinates of Point \(A\)**: Since \(O\) and \(B\) are two vertices of the square, the coordinates of point \(A\) can be found using the fact that \(OA\) is perpendicular to \(OB\) and has the same length. The slope of line \(OB\) is: \[ \text{slope of } OB = \frac{-6 - 0}{8 - 0} = -\frac{3}{4} \] The slope of line \(OA\) will be the negative reciprocal of the slope of \(OB\): \[ \text{slope of } OA = \frac{4}{3} \] 4. **Finding the Equation of Line \(OA\)**: Using point-slope form, the equation of line \(OA\) through point \(O(0,0)\) is: \[ y = \frac{4}{3}x \] 5. **Finding the Coordinates of Point \(A\)**: Since \(OA\) is equal in length to \(OB\) (which is 10), we can set up the equation: \[ \sqrt{(a - 0)^2 + (b - 0)^2} = 10 \] This simplifies to: \[ a^2 + b^2 = 100 \] 6. **Finding the Coordinates of Point \(C\)**: The coordinates of point \(C\) can be found using the midpoint of \(OB\) and the fact that \(C\) is also a vertex of the square. The midpoint \(M\) of \(OB\) is: \[ M = \left(\frac{0 + 8}{2}, \frac{0 - 6}{2}\right) = (4, -3) \] The coordinates of point \(C\) can be determined by maintaining the perpendicular distance from \(M\) to \(C\) equal to the distance from \(M\) to \(A\). 7. **Finding \(|a| + |b|\)**: Since we have \(a^2 + b^2 = 100\) and \(A\) lies on the line \(y = \frac{4}{3}x\), we can substitute \(b = \frac{4}{3}a\) into the equation: \[ a^2 + \left(\frac{4}{3}a\right)^2 = 100 \] This leads to: \[ a^2 + \frac{16}{9}a^2 = 100 \implies \frac{25}{9}a^2 = 100 \implies a^2 = 36 \implies a = 6 \text{ or } -6 \] Substituting \(a = 6\) into \(b = \frac{4}{3}a\): \[ b = \frac{4}{3} \cdot 6 = 8 \] Thus, \(|a| + |b| = |6| + |8| = 6 + 8 = 14\). ### Final Answer: \[ |a| + |b| = 14 \]
Promotional Banner

Topper's Solved these Questions

  • COMPLEX NUMBERS

    VMC MODULES ENGLISH|Exercise JEE ARCHIVE|76 Videos
  • COMPLEX NUMBERS

    VMC MODULES ENGLISH|Exercise LEVEL - 2|50 Videos
  • CIRCLES

    VMC MODULES ENGLISH|Exercise JEE ADVANCED ( ARCHIVE )|68 Videos
  • CONIC SECTIONS

    VMC MODULES ENGLISH|Exercise JEE ADVANCED ARCHIVE|76 Videos

Similar Questions

Explore conceptually related problems

A square has one vertex at the vertex of the parabola y^2=4a x and the diagonal through the vertex lies along the axis of the parabola. If the ends of the other diagonal lie on the parabola, the coordinates of the vertices of the square are (a) (4a ,4a) (b) (4a ,-4a) (c) (0,0) (d) (8a ,0)

Three vertices of a parallelogram ABCD are B (6,7) , C (8,3) and D (0,-1) . Find the co-ordinates of vertex A .

If A and B are the points (-3,4) and (3,-4) respectively then the co-ordinates of the point C on AB produced such that AC=3BC are (A) (0,0) (B) (-6,8) (C) (6,-8) (D) (-6,-8)

The value of det |(a,0,0,0),(2,b,0,0),(4,6,c,0),(6,8,10,d)| is (A) 0 (B) a+b+c+d (C) abcd (D) none of these

If A = {2,4,6,8} , B = {3,6,9,12} and C = {0,5,10,15} then the overlapping sets are (a) A and B (b) B and C (c) A and C (d) none of these

If A,B,C are three square matrices of the same order, then A B=A C => B=Cdot Then |A|!=0 b. A is invertible c. A may be orthogonal d. is symmetric

The vertex of the parabola (y+a)^2=8a(x-a) is a. (-a ,-a) b. (a ,-a) c. (-a , a) d. none of these

If the points (a, 0), (b,0), (0, c) , and (0, d) are concyclic (a, b, c, d > 0) , then prove that ab = cd .

If (-1,\ 2),\ (2,\ -1) and (3,\ 1) are any three vertices of a parallelogram then the fourth vertex (a,b) will be such that (a) a=2 , b=0 (b) a= - 2, b=0 (c) a= - 2, b=6 (d) a=0, b=4

y^(2)=4xandy^(2)=-8(x-a) intersect at points A and C. Points O (0,0) , A,B (a,0) , and C are concyclic. The area of cyclic quadrilateral OABC is