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Find the equation of the straight line through P and Q, the coordinates of P and Q, and calculate the value of the shaded area.

Mathematics

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3 parts
a(i)

y = -x + 7\text{y = -x + 7}

a(ii)

P(-1, 8) and Q(2, 5)\text{P(-1, 8) and Q(2, 5)}

a(iii)

4.5 square units\text{4.5 square units}

Step-by-step solution

10 steps

Here's the solution for question 6:

Part a(i)· 2 steps

Write down the equation of the straight line through P and Q.

  1. Find the gradient mm of the line passing through (0,7)(0, 7) and (7,0)(7, 0).

    m=y2−y1x2−x1=0−77−0=−77=−1m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - 7}{7 - 0} = \frac{-7}{7} = -1

  2. Use the point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1) with the point (0,7)(0, 7).

    y−7=−1(x−0)y - 7 = -1(x - 0) y−7=−xy - 7 = -x y=−x+7y = -x + 7

    The equation of the straight line is:

    y = -x + 7\boxed{\text{y = -x + 7}}

Part a(ii)

P(-1, 8) and Q(2, 5)\text{P(-1, 8) and Q(2, 5)}

Find the coordinates of P and Q.

  1. Set the equation of the line equal to the equation of the curve to find the intersection points.

    The curve is y=(x−1)2+4y = (x-1)^2 + 4.
    The line is y=−x+7y = -x + 7.

    −x+7=(x−1)2+4-x + 7 = (x-1)^2 + 4

  2. Expand and simplify the equation.

    −x+7=x2−2x+1+4-x + 7 = x^2 - 2x + 1 + 4 −x+7=x2−2x+5-x + 7 = x^2 - 2x + 5 0=x2−2x+x+5−70 = x^2 - 2x + x + 5 - 7 0=x2−x−20 = x^2 - x - 2

  3. Factor the quadratic equation to find the x-coordinates.

    (x−2)(x+1)=0(x - 2)(x + 1) = 0

    So, x=2x = 2 or x=−1x = -1.

  4. Substitute the x-values back into the line equation y=−x+7y = -x + 7 to find the corresponding y-coordinates.

    For x=2x = 2:

    y=−(2)+7=5y = -(2) + 7 = 5

    Point Q is (2,5)(2, 5).
    For x=−1x = -1:

    y=−(−1)+7=1+7=8y = -(-1) + 7 = 1 + 7 = 8

    Point P is (−1,8)(-1, 8).
    The coordinates of P and Q are:

    P(-1, 8) and Q(2, 5)\boxed{\text{P(-1, 8) and Q(2, 5)}}

Part a(iii)

4.5 square units\text{4.5 square units}

Calculate the value of the shaded area.

  1. The shaded area is the area between the line yL=−x+7y_L = -x + 7 and the curve yC=(x−1)2+4=x2−2x+5y_C = (x-1)^2 + 4 = x^2 - 2x + 5 from x=−1x = -1 to x=2x = 2.

    The line is above the curve in this region.

    Area=∫−12(yL−yC) dx\text{Area} = \int_{-1}^{2} (y_L - y_C)\,dx Area=∫−12((−x+7)−(x2−2x+5)) dx\text{Area} = \int_{-1}^{2} ((-x + 7) - (x^2 - 2x + 5))\,dx

  2. Simplify the integrand.

    Area=∫−12(−x+7−x2+2x−5) dx\text{Area} = \int_{-1}^{2} (-x + 7 - x^2 + 2x - 5)\,dx Area=∫−12(−x2+x+2) dx\text{Area} = \int_{-1}^{2} (-x^2 + x + 2)\,dx

  3. Integrate the expression.

    Area=[−x33+x22+2x]−12\text{Area} = \left[-\frac{x^3}{3} + \frac{x^2}{2} + 2x\right]_{-1}^{2}

  4. Evaluate the definite integral.

    Area=(−(2)33+(2)22+2(2))−(−(−1)33+(−1)22+2(−1))\text{Area} = \left(-\frac{(2)^3}{3} + \frac{(2)^2}{2} + 2(2)\right) - \left(-\frac{(-1)^3}{3} + \frac{(-1)^2}{2} + 2(-1)\right) Area=(−83+42+4)−(−−13+12−2)\text{Area} = \left(-\frac{8}{3} + \frac{4}{2} + 4\right) - \left(-\frac{-1}{3} + \frac{1}{2} - 2\right) Area=(−83+2+4)−(13+12−2)\text{Area} = \left(-\frac{8}{3} + 2 + 4\right) - \left(\frac{1}{3} + \frac{1}{2} - 2\right) Area=(−83+6)−(13+12−42)\text{Area} = \left(-\frac{8}{3} + 6\right) - \left(\frac{1}{3} + \frac{1}{2} - \frac{4}{2}\right) Area=(−8+183)−(2+3−126)\text{Area} = \left(\frac{-8 + 18}{3}\right) - \left(\frac{2 + 3 - 12}{6}\right) Area=103−(−76)\text{Area} = \frac{10}{3} - \left(\frac{-7}{6}\right) Area=103+76\text{Area} = \frac{10}{3} + \frac{7}{6} Area=206+76\text{Area} = \frac{20}{6} + \frac{7}{6} Area=276=92=4.5\text{Area} = \frac{27}{6} = \frac{9}{2} = 4.5

    The value of the shaded area is:

    4.5 square units\boxed{\text{4.5 square units}}
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