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2 The diagram shows the points 
A(0,-1),B(p,p) and 
C(p,q).
(a) Given that the length of 
AB is 5 units, find the value of 
p.

22 The diagram shows the points A(0,1),B(p,p) A(0,-1), B(p, p) and C(p,q) C(p, q) .\newline(a) Given that the length of AB A B is 55 units, find the value of p p .

Full solution

Q. 22 The diagram shows the points A(0,1),B(p,p) A(0,-1), B(p, p) and C(p,q) C(p, q) .\newline(a) Given that the length of AB A B is 55 units, find the value of p p .
  1. Calculate distance formula: Calculate the distance between points A(0,1)A(0,-1) and B(p,p)B(p,p) using the distance formula: d=((x2x1)2+(y2y1)2)d = \sqrt{((x_2 - x_1)^2 + (y_2 - y_1)^2)}.
  2. Substitute coordinates into formula: Substitute the coordinates of AA and BB into the distance formula: d=((p0)2+(p(1))2)d = \sqrt{((p - 0)^2 + (p - (-1))^2)}.
  3. Simplify expression: Simplify the expression: d=p2+(p+1)2d = \sqrt{p^2 + (p + 1)^2}.
  4. Set distance equal to 55: Since ABAB is 55 units, set the distance equal to 55: 5=p2+(p+1)25 = \sqrt{p^2 + (p + 1)^2}.
  5. Square both sides: Square both sides to remove the square root: 25=p2+(p+1)225 = p^2 + (p + 1)^2.
  6. Expand squared term: Expand the squared term: 25=p2+p2+2p+125 = p^2 + p^2 + 2p + 1.
  7. Combine like terms: Combine like terms: 25=2p2+2p+125 = 2p^2 + 2p + 1.
  8. Set equation to zero: Subtract 2525 from both sides to set the equation to zero: 0=2p2+2p240 = 2p^2 + 2p - 24.
  9. Divide entire equation: Divide the entire equation by 22 to simplify: 0=p2+p120 = p^2 + p - 12.
  10. Factor quadratic equation: Factor the quadratic equation: p + \(4)(p - 33) = 00\
  11. Solve for pp: Set each factor equal to zero and solve for pp: p+4=0p + 4 = 0 or p3=0p - 3 = 0.
  12. Final solution: Solve for pp: p=4p = -4 or p=3p = 3.
  13. Final solution: Solve for pp: p=4p = -4 or p=3p = 3.Since pp is a coordinate, it cannot be negative in this context; therefore, p=3p = 3.

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