Find the distance between the line and the point in each of the following :
`(i) 3x+4y-5=0`, `(-3,4)`
`(ii) 12x-5y-7=0`, `(3,-1)`
`(iii) 3x-4y-26=0`, `(3,-5)`
`(iv) x+y=0`, `(0,0)`
`(v)y=4` , `(2,3)`.

Answer

To find the distance between a line and a point, we can use the formula for the perpendicular distance from a point \((x_1, y_1)\) to a line given by the equation \(Ax + By + C = 0\):

\[ D = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]

where \(D\) is the distance, and \(A\), \(B\), and \(C\) are the coefficients from the line equation.

Now, let's solve the questions step by step.

### (i) For the line \(3x + 4y - 5 = 0\) and the point \((-3, 4)\):

1. Identify \(A\), \(B\), and \(C\):
- \(A = 3\), \(B = 4\), \(C = -5\)

2. Substitute the point \((-3, 4)\) into the distance formula:
\[ D = \frac{|3(-3) + 4(4) - 5|}{\sqrt{3^2 + 4^2}} \]

3. Calculate the numerator:
\[ = | -9 + 16 - 5 | = | 2 | = 2 \]

4. Calculate the denominator:
\[ = \sqrt{9 + 16} = \sqrt{25} = 5 \]

5. Finally, compute the distance:
\[ D = \frac{2}{5} \]

### (ii) For the line \(12x - 5y - 7 = 0\) and the point \((3, -1)\):

1. Identify \(A\), \(B\), and \(C\):
- \(A = 12\), \(B = -5\), \(C = -7\)

2. Substitute the point \((3, -1)\) into the distance formula:
\[ D = \frac{|12(3) - 5(-1) - 7|}{\sqrt{12^2 + (-5)^2}} \]

3. Calculate the numerator:
\[ = |36 + 5 - 7| = |34| = 34 \]

4. Calculate the denominator:
\[ = \sqrt{144 + 25} = \sqrt{169} = 13 \]

5. Finally, compute the distance:
\[ D = \frac{34}{13} \]

### (iii) For the line \(3x - 4y - 26 = 0\) and the point \((3, -5)\):

1. Identify \(A\), \(B\), and \(C\):
- \(A = 3\), \(B = -4\), \(C = -26\)

2. Substitute the point \((3, -5)\) into the distance formula:
\[ D = \frac{|3(3) - 4(-5) - 26|}{\sqrt{3^2 + (-4)^2}} \]

3. Calculate the numerator:
\[ = |9 + 20 - 26| = |3| = 3 \]

4. Calculate the denominator:
\[ = \sqrt{9 + 16} = \sqrt{25} = 5 \]

5. Finally, compute the distance:
\[ D = \frac{3}{5} \]

### (iv) For the line \(x + y = 0\) and the point \((0, 0)\):

1. Identify \(A\), \(B\), and \(C\):
- \(A = 1\), \(B = 1\), \(C = 0\)

2. Substitute the point \((0, 0)\) into the distance formula:
\[ D = \frac{|1(0) + 1(0) + 0|}{\sqrt{1^2 + 1^2}} \]

3. Calculate the numerator:
\[ = |0| = 0 \]

4. Calculate the denominator:
\[ = \sqrt{1 + 1} = \sqrt{2} \]

5. Finally, compute the distance:
\[ D = \frac{0}{\sqrt{2}} = 0 \]

### (v) For the line \(y = 4\) and the point \((2, 3)\):

1. Identify \(A\), \(B\), and \(C\):
- The line can be rewritten as \(0x + 1y - 4 = 0\), so \(A = 0\), \(B = 1\), \(C = -4\)

2. Substitute the point \((2, 3)\) into the distance formula:
\[ D = \frac{|0(2) + 1(3) - 4|}{\sqrt{0^2 + 1^2}} \]

3. Calculate the numerator:
\[ = |3 - 4| = |-1| = 1 \]

4. Calculate the denominator:
\[ = \sqrt{0 + 1} = 1 \]

5. Finally, compute the distance:
\[ D = \frac{1}{1} = 1 \]

### Summary of Distances:
1. \(D = \frac{2}{5}\)
2. \(D = \frac{34}{13}\)
3. \(D = \frac{3}{5}\)
4. \(D = 0\)
5. \(D = 1\)