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1 Determine the co-ordinate of M the midpoint of line RT. 3.2 Calculate the gradient of the line WT. 3.3 Prove that RW is perpendicular to WT.

Mathematics

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3 parts
3.1

(12,92)\left(\frac{1}{2}, \frac{9}{2}\right)

3.2

23\frac{2}{3}

3.4

a=4a = 4

b=9b = 9

Step-by-step solution

9 steps

Here are the solutions to Question 3:

Part 3.1· 2 steps

Determine the co-ordinate of M the midpoint of line RT.

To find the midpoint M of a line segment with endpoints (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2), we use the formula:

M=(x1+x22,y1+y22)M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)

Given R(−5,3-5, 3) and T(6,66, 6).

  1. Substitute the coordinates of R and T into the midpoint formula.

    M=(−5+62,3+62)M = \left(\frac{-5 + 6}{2}, \frac{3 + 6}{2}\right)
  2. Calculate the coordinates of M.

    M=(12,92)M = \left(\frac{1}{2}, \frac{9}{2}\right)

    The coordinates of M are (12,92)\boxed{\left(\frac{1}{2}, \frac{9}{2}\right)}.

Part 3.2

23\frac{2}{3}

Calculate the gradient of the line WT.

To calculate the gradient mm of a line passing through two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2), we use the formula:

m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}

Given W(−3,0-3, 0) and T(6,66, 6).

  1. Substitute the coordinates of W and T into the gradient formula.

    mWT=6−06−(−3)m_{WT} = \frac{6 - 0}{6 - (-3)}
  2. Simplify the expression to find the gradient.

    mWT=66+3m_{WT} = \frac{6}{6 + 3} mWT=69m_{WT} = \frac{6}{9} mWT=23m_{WT} = \frac{2}{3}

    The gradient of line WT is 23\boxed{\frac{2}{3}}.

Part 3.3· 2 steps

Prove that RW is perpendicular to WT.

For two lines to be perpendicular, the product of their gradients must be −1-1. We already have mWT=23m_{WT} = \frac{2}{3}. Now we need to calculate the gradient of RW.
Given R(−5,3-5, 3) and W(−3,0-3, 0).

  1. Calculate the gradient of RW.

    mRW=0−3−3−(−5)m_{RW} = \frac{0 - 3}{-3 - (-5)} mRW=−3−3+5m_{RW} = \frac{-3}{-3 + 5} mRW=−32m_{RW} = \frac{-3}{2}
  2. Multiply the gradients of RW and WT.

    mRW×mWT=(−32)×(23)m_{RW} \times m_{WT} = \left(-\frac{3}{2}\right) \times \left(\frac{2}{3}\right) mRW×mWT=−1m_{RW} \times m_{WT} = -1

    Since the product of their gradients is −1-1, RW is indeed perpendicular to WT.

Part 3.4

a=4a = 4

If RWTK is a parallelogram, determine the values of a and b.

In a parallelogram, the diagonals bisect each other, meaning they share a common midpoint. The diagonals of RWTK are RT and WK. We already found the midpoint of RT in 3.1 as MRT=(12,92)M_{RT} = \left(\frac{1}{2}, \frac{9}{2}\right).
Now we find the midpoint of WK using W(−3,0-3, 0) and K(a,ba, b).

  1. Set up the midpoint formula for WK.

    MWK=(−3+a2,0+b2)M_{WK} = \left(\frac{-3 + a}{2}, \frac{0 + b}{2}\right)
  2. Equate the x-coordinates of MRTM_{RT} and MWKM_{WK}.

    −3+a2=12\frac{-3 + a}{2} = \frac{1}{2} −3+a=1-3 + a = 1 a=1+3a = 1 + 3 a=4a = 4
  3. Equate the y-coordinates of MRTM_{RT} and MWKM_{WK}.

    0+b2=92\frac{0 + b}{2} = \frac{9}{2} b=9b = 9

    The values are a=4\boxed{a = 4} and b=9\boxed{b = 9}.

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