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​

Answer 1

4 units

Explanation:

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

`\displaystyle{d = \fracax_1 + by_1 + c { \sqrt{ {a}^(2) + {b}^(2) } }}`

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

`\displaystyle{d = \frac { \sqrt{ {3}^(2) + {4}^(2) } }}`

Then evaluate the value:

`\displaystyle 9+ 16 - 5 \\ \\ \displaystyle 20 \\ \\ \displaystyle{d = (20)/( 5)} \\ \\ \displaystyle{d = 4 \: \text{units}}`

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