Mathematics

Question

12. The distance from the point p(3,4) to the line L with equation: 3x + 4y = 5 is A. 6 units B. 7units C. 4 units D. 5 units of the​

1 Answer

  • Answer:

    4 units

    Step-by-step explanation:

    The distance between P(x1,y1) and the line is:

    [tex] \displaystyle{d = \dfrac{ |ax_1 + by_1 + c | }{ \sqrt{ {a}^{2} + {b}^{2} } }}[/tex]

    From the line equation 3x + 4y = 5 can be arranged as 3x + 4y - 5 = 0 with P(3,4). Therefore, we will have:

    [tex] \displaystyle{d = \dfrac{ |3(3) + 4(4) - 5| }{ \sqrt{ {3}^{2} + {4}^{2} } }}[/tex]

    Then evaluate the value:

    [tex] \displaystyle{d = \dfrac{ | 9+ 16 - 5| }{ \sqrt{ 9 + 16}}} \\ \\ \displaystyle{d = \dfrac{ | 20| }{ \sqrt{ 25}}} \\ \\ \displaystyle{d = \dfrac{20}{ 5}} \\ \\ \displaystyle{d = 4 \: \text{units}}[/tex]

    Therefore, the distance between point (3,4) and the line is 4 units.

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