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
Mathematics
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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

1. User Answers VectorFundament120
Answer:
4 units
Stepbystep 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.