Question

( Exercises 11.8- Complete the following: 1. Find the distance between the line and the point. (a) 3x + 4y = 24; (-10, 1) (c) x + 2y = 12; (4, -6) (e) x + 3y = 0; (1, -7)

Answer 1

Answer : Distance between the equation and the point is 7.07

We are given the equation of a line to be

x + 3y = 0

Firstly, we need to find the value of x and y from the equation

To find x, isolate y by making it zero

x + 3(0) = 0

x + 0 = 0

x = 0 - 0

x = 0

To find y, make x = 0

0 + 3y = 0

3y = 0

Divide both sides by 3

3y/3 = 0/3

y = 0

Therefore, x = 0, and y = 0

(0 , 0) and (1, -7)

From the given points

x1 = 0, y1 = 0, x2 = 1, y2 = -7

`\begin{gathered} \text{Distance = }\sqrt[]{(x1-x2)^2+(y1-y2)^2} \\ \text{Distance = }\sqrt[]{(0-1)^2+(0-(-7)\rbrack^2} \\ \text{Distance = }\sqrt[]{(-1)^2+(0+7)^2} \\ \text{Distance = }\sqrt[]{(-1)^2+(7)^2} \\ \text{Distance = }\sqrt[]{1\text{ + 49}} \\ \text{Distance = }\sqrt[]{50} \\ \text{Distance = 7.07 units} \end{gathered}`