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Math Help - coordinate geometry of straight lines

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    coordinate geometry of straight lines

    The point P has coordinates (1,10) and the point Q has coordinates (4,4).

    (a) Show the length of PQ is 3 squared 5.

    (b) (i) Find the equation of the perpendicular bisector of PQ.
    (ii) This perpendicular bisector intersects the x-axis at the point A. Find the coordinates of A.
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    Quote Originally Posted by Sam16 View Post
    The point P has coordinates (1,10) and the point Q has coordinates (4,4).

    (a) Show the length of PQ is 3 squared 5.
    PQ =\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

    PQ =\sqrt{(4-1)^2+(4-10)^2}

    PQ =\sqrt{(3)^2+(-6)^2}

    PQ =\sqrt{9+36}

    PQ =\sqrt{45}=\sqrt{9 \cdot 5}=\sqrt{9} \cdot \sqrt{5}=3\sqrt{5}
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    Quote Originally Posted by Sam16 View Post
    The point P has coordinates (1,10) and the point Q has coordinates (4,4).

    (a) Show the length of PQ is 3 squared 5.

    (b) (i) Find the equation of the perpendicular bisector of PQ.
    (ii) This perpendicular bisector intersects the x-axis at the point A. Find the coordinates of A.
    to (a): Use the distance formula

    d(PQ)=\sqrt{(x_P-x_Q)^2+(y_P -y_Q)^2}

    to (b)(i)
    1. Calculate the slope m_{PQ}
    2. Calculate the perpendicular direction to this slope
    3. Calculate the coordinates of the mid point of PQ: M_{PQ}\left(\frac{x_P+x_Q}2~,~\frac{y_P+y_Q}2\righ  t)
    4. Use the point-slope-formula to get the equation of the perpendicular bisector.

    to (b)(ii)
    Solve the equation y = 0 using the equation of (b)(i)4
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    Quote Originally Posted by Sam16 View Post
    The point P has coordinates (1,10) and the point Q has coordinates (4,4).

    (b) (i) Find the equation of the perpendicular bisector of PQ.
    (ii) This perpendicular bisector intersects the x-axis at the point A. Find the coordinates of A.
    Find the midpoint of PQ.

    Midpoint = \left(\frac{1+4}{2} \ \ , \ \ \frac{10+4}{2}\right) = \left(\frac{5}{2} \ \ , \ \ 7\right)

    Find the slop of PQ.

    m=\frac{4-10}{4-1}=-\frac{6}{3}=-2

    The slope of the perpendicular will be the negative reciprocal of that slope which is \frac{1}{2}

    The equation of the perpendicular will have slope = 1/2 and pass through the midpoint of PQ which is (5/2, 7).

    y = mx + b

    7=\frac{1}{2}\left(\frac{5}{2}\right)+b

    7=\frac{5}{4}+b

    \frac{23}{4}=b

    y=\frac{1}{2}x+\frac{23}{4}

    The above is the equation of the perpendicular in slope-intercept form.

    General form would be 2x-4y+23=0

    To find the x-intercept, simply assign 0 to y and solve for x. The coordinates will be in the form (x, 0).
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