Question

The two lines y = 2x + 8 and y = 2x – 12 intersect the x-axis at the P and Q.
Work out the distance PQ.

Answers

  1. Answer:
    PQ = 10 units
    Step-by-step explanation:
    to find where the lines cross the x- axis let y = 0 and solve for x , that is
    2x + 8 = 0 ( subtract 8 from both sides )
    2x = – 8 ( divide both sides by 2 )
    x = – 4 ← point P
    and
    2x – 12 = 0 ( add 12 to both sides )
    2x = 12 ( divide both sides by 2 )
    x = 6 ← point Q
    the lines cross the x- axis at x = – 4 and x = 6
    using the absolute value of the difference , then
    PQ = | – 4 – 6 | = | – 10 | = 10 units
    or
    PQ = | 6 – (- 4) | = | 6 + 4 | = | 10 | = 10 units

    Reply
  2. Answer: \Huge\boxed{Distance=10~units}
    Step-by-step explanation:

    Find the point P

    Given expression
    y = 2x + 8
    Substitute 0 for the y value to find the x value
    This is the definition of x-intercepts
    (0) = 2x + 8
    Subtract 8 on both sides
    0 – 8 = 2x + 8 – 8
    -8 = 2x
    Divide 2 on both sides
    -8 / 2 = 2x / 2
    x = -4
    \large\boxed{P~(-4,0)}

    Find the point Q

    Given expression
    y = 2x – 12
    Substitute 0 for the y value to find the x value
    (0) = 2x – 12
    Add 12 on both sides
    0 + 12 = 2x – 12 + 12
    12 = 2x
    Divide 2 on both sides
    12 / 2 = 2x / 2
    x = 6
    \large\boxed{Q~(6,0)}

    Find the distance between PQ

    Given information
    (x_1,~y_1)=(-4,~0)
    (x_2,~y_2)=(6,~0)
    Substitute values into the distance formula
    Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}
    Distance=\sqrt{(6-(-4))^2+(0-0)^2}
    Simplify values in the parenthesis
    Distance=\sqrt{(10)^2+(0)^2}
    Simplify values in the radical sign
    Distance=\sqrt{100}
    \Huge\boxed{Distance=10~units}
    Hope this helps!! 🙂
    Please let me know if you have any questions

    Reply

Leave a Comment