Anonymous
Anonymous asked in Science & MathematicsMathematics · 7 years ago

How to find values of y when 3 coordinates are given?

Can someone help me answer this question. Would really appreciate it :)

Three points A, B and C have co-ordinates (1,3), (3,5) and (-1,y). Find the values of y when:

i) AB = AC

ii) AC = BC

iii) AS is perpendicular to BC

iv) A, B and C are collinear

Please can u include all steps

Many thanks :)

2 Answers

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    We know that the distance between AB and AC is the same

    AB =>>

    d^2 = (5 - 3)^2 + (3 - 1)^2

    d^2 = 2^2 + 2^2

    d^2 = 2 * 2^2

    d = 2 * sqrt(2)

    AC =>>

    (2 * sqrt(2))^2 = (y - 3)^2 + (-1 - 1)^2

    8 = (y - 3)^2 + (-2)^2

    8 = (y - 3)^2 + 4

    4 = (y - 3)^2

    +/- 2 = y - 3

    y = 3 +/- 2

    y = 5 , 1

    We have 2 possible values for y

    ii)

    AC and BC are the same

    (y - 3)^2 + (-1 - 1)^2 = (y - 5)^2 + (-1 - 3)^2

    y^2 - 6y + 9 + 4 = y^2 - 10y + 25 + 16

    -6y + 13 = -10y + 41

    -6y + 10y = 41 - 13

    4y = 28

    y = 7

    iii)

    AS? What's S? Did you mean AC?

    Slope for AC

    (y - 3) / (-1 - 1) =>

    (y - 3) / (-2) =>

    (3 - y) / 2

    Slope for BC

    (y - 5) / (-1 - 3) =>

    (y - 5) / (-4) =>

    (5 - y) / 4

    mAC = -1 / mBC =>>

    (3 - y) / 2 = -4 / (5 - y)

    (3 - y) * (5 - y) = -8

    15 - 5y - 3y + y^2 = -8

    15 - 8y + y^2 = -8

    y^2 - 8y + 16 = -8 + 1

    (y - 4)^2 = -7

    y - 4 = +/- i * sqrt(7)

    y = 4 +/- i * sqrt(7)

    Unless you meant that AB is perpendicular to BC

    (5 - 3) / (3 - 1) => 2/2 = 1

    (y - 5) / (-1 - 3) = -1 / (1)

    (y - 5) / (-4) = -1

    y - 5 = 4

    y = 9

    iv)

    A, B, and C are collinear

    Find the slope between A and B

    (5 - 3) / (3 - 1) = 2/2 = 1

    y - 3 = 1 * (x - 1)

    y - 3 = x - 1

    y = x + 2

    x = -1

    y = -1 + 2

    y = 1

  • Hosam
    Lv 6
    7 years ago

    i) If AB = AC , then

    (1-3)^2 + (3 - 5)^2 = (1 - (-1))^2 + (y - 3)^2

    8 = 4 + (y - 3)^2

    (y - 3)^2 = 4

    y = 5, or y = 1.

    ii) AC = BC ==>

    (2)^2 + (y-3)^2 = (4)^2 + (y - 5)^2

    4 + y^2 - 6 y + 9 = 16 + y^2 - 10 y + 25

    4 y = 28

    y = 7

    iii) AC is perpendicular to BC.

    AC = (-2, y-3)

    BC = (-4, y-5)

    AC . BC = 0 ==> 8 + (y-3)(y-5) = 0 ==> y^2 - 8 y + 23 = 0

    Discriminant 8^2 - 4*23 = 64 - 92 < 0

    So there is no value of y that will satisfy the condition.

    iv) If the three points are collinear, then (5 - 3)/(3 - 1) = (y - 3)/(-1 - 1)

    Therefore y - 3 = -2, and y = 1.

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