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Math Help - Finding something in co-ordinate geometery

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    Finding something in co-ordinate geometery

    The question said that the points of A, B and C were (-2,-3), (6,1) and (k,3) respectively, and that AB is perpendicular to BC.

    It asked to work out the gradient of AB:

    y2-y1 / x2-x1 = 4/8 --> 1/2.

    I think that might be right but then it says to prove that the value of k equals 5. How could I do that?
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    A riddle wrapped in an enigma
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    Quote Originally Posted by db5vry View Post
    The question said that the points of A, B and C were (-2,-3), (6,1) and (k,3) respectively, and that AB is perpendicular to BC.

    It asked to work out the gradient of AB:

    y2-y1 / x2-x1 = 4/8 --> 1/2.

    I think that might be right but then it says to prove that the value of k equals 5. How could I do that?
    A(-2, -3)
    B(6, 1)
    C(k, 3)

    Your slope of AB is correct. Since AB is perpendicular to BC, this means that the slope of BC must be the negative reciprocal of the slope of AB.

    Therefore, the slope of BC must be -2.

    Solve this: \frac{3-1}{k-6}=-2 for k.
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    If AB is perpendicular to BC then the dot product is equal to 0.
    AB = (6,1)-(-2,-3)=(8,4) and BC = (k,3) - (6,1) = (k-6,2)
    *AB \cdot BC =  (8,4)\cdot (k-6,2) =8k - 48 +8 =0 \implies k =5
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    Quote Originally Posted by masters View Post
    A(-2, -3)
    B(6, 1)
    C(k, 3)

    Your slope of AB is correct. Since AB is perpendicular to BC, this means that the slope of BC must be the negative reciprocal of the slope of AB.

    Therefore, the slope of BC must be -2.

    Solve this: \frac{3-1}{k-6}=-2 for k.
    Thanks for that. It was really helpful.

    The next question says that the line L is parallel to BC and passes through A, asking you to figure out an equation for L:

    I used the formula "y-y1=m(x-x1)", A(-2,-3) and the -2 gradient from earlier used to put into values as this:

    y--3 = -2(x--2)
    y+3 = -2(x+2)
    y+3 = -2x-4
    y+2x-7=0 [L]

    Is this right? Thanks again for your help.
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    A riddle wrapped in an enigma
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    Quote Originally Posted by db5vry View Post
    Thanks for that. It was really helpful.

    The next question says that the line L is parallel to BC and passes through A, asking you to figure out an equation for L:

    I used the formula "y-y1=m(x-x1)", A(-2,-3) and the -2 gradient from earlier used to put into values as this:

    y--3 = -2(x--2)
    y+3 = -2(x+2)
    y+3 = -2x-4
    y+2x+7=0 [L]

    Is this right? Thanks again for your help.
    One change. See sign change in red above. And, you might want to put the equation in one of the more recognizable forms:

    Standard Form: 2x+y=-7
    Slope-Intercept Form: y=-2x-7
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    Quote Originally Posted by masters View Post
    One change. See sign change in red above. And, you might want to put the equation in one of the more recognizable forms:

    Standard Form: 2x+y=-7
    Slope-Intercept Form: y=-2x-7
    I'll use the slope-intercept form there then. It makes more sense! And I see how you get +7 because you need to take the 4 over to the other side, I didn't see that.

    There's only one more part to the question, it says that the line intersects the y-axis at D and asks to calculate the length of CD.

    Now here, I'm very stuck and not sure what formula to use. Can you help at all? If you can thanks again!
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    A riddle wrapped in an enigma
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    Quote Originally Posted by db5vry View Post
    I'll use the slope-intercept form there then. It makes more sense! And I see how you get +7 because you need to take the 4 over to the other side, I didn't see that.

    There's only one more part to the question, it says that the line intersects the y-axis at D and asks to calculate the length of CD.

    Now here, I'm very stuck and not sure what formula to use. Can you help at all? If you can thanks again!
    The y-intercept of this line is (0, -7).

    Use the distance formula to find the distance from C (5, 3) to D (0, -7)

    d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}
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