To solve the problem step by step, we will follow the outlined approach in the video transcript.
### Step 1: Find the intersection point of the lines
We have two lines given by the equations:
1. \( 4x + y - 1 = 0 \) (Equation 1)
2. \( 7x - 3y - 35 = 0 \) (Equation 2)
First, we can express \( y \) from Equation 1:
\[
y = 1 - 4x
\]
Now, substitute this expression for \( y \) into Equation 2:
\[
7x - 3(1 - 4x) - 35 = 0
\]
Expanding this gives:
\[
7x - 3 + 12x - 35 = 0
\]
Combining like terms:
\[
19x - 38 = 0
\]
Solving for \( x \):
\[
x = 2
\]
Now substitute \( x = 2 \) back into the expression for \( y \):
\[
y = 1 - 4(2) = 1 - 8 = -7
\]
Thus, the intersection point of the two lines is \( (2, -7) \).
### Step 2: Find the equation of the line joining the points \( (3, 5) \) and \( (2, -7) \)
Let \( (x_1, y_1) = (3, 5) \) and \( (x_2, y_2) = (2, -7) \).
The slope \( m \) of the line joining these two points is given by:
\[
m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-7 - 5}{2 - 3} = \frac{-12}{-1} = 12
\]
Using the point-slope form of the equation of a line:
\[
y - y_1 = m(x - x_1)
\]
Substituting the values:
\[
y - 5 = 12(x - 3)
\]
Expanding this:
\[
y - 5 = 12x - 36
\]
Rearranging gives:
\[
12x - y - 31 = 0
\]
### Step 3: Check if the line is equidistant from the points \( (0, 0) \) and \( (8, 34) \)
Using the formula for the distance from a point \( (x_0, y_0) \) to a line \( Ax + By + C = 0 \):
\[
d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}
\]
For the line \( 12x - y - 31 = 0 \):
- \( A = 12 \)
- \( B = -1 \)
- \( C = -31 \)
**Distance from point \( (0, 0) \)**:
\[
d_1 = \frac{|12(0) - 1(0) - 31|}{\sqrt{12^2 + (-1)^2}} = \frac{31}{\sqrt{144 + 1}} = \frac{31}{\sqrt{145}}
\]
**Distance from point \( (8, 34) \)**:
\[
d_2 = \frac{|12(8) - 1(34) - 31|}{\sqrt{12^2 + (-1)^2}} = \frac{|96 - 34 - 31|}{\sqrt{145}} = \frac{|31|}{\sqrt{145}} = \frac{31}{\sqrt{145}}
\]
Since \( d_1 = d_2 \), the line is equidistant from the points \( (0, 0) \) and \( (8, 34) \).
### Conclusion
Thus, the statement is **True**.
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