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.
