Given, A line segment joining (4, 2) and (3, 5) if it cuts off an intercept 3 from y–axis.

To Find: The equation of that line. 

Formula used: The equation of line is y = mx + C 

Explanation: Here, The required equation of line is y = mx + c 

Now, c = 3 (Given)

Let m be slope of given line = – 1

Slope of line joining (x₁ – x₂) and (y₁ – y₂) ,m = \(\frac{y_2-y_1}{x_1-x_2}\)

So, Slope of line joining (4, 2) and (3, 5),m = \(\frac{5-2}{3-4}\) = \(\frac{2}{-1}\) 

Therefore, m = \(\frac{1}{3}\)

Now, The equation of line is y = mx + c

y = \(\frac{1}{3}x+3\) 

x – 3y + 9 = 0

Hence, The equation of line is 2y + 5x + 6 = 0.

Given: 

The equation is perpendicular to √3x – y + 5 = 0 equation and cuts off an intercept of 4 units with the negative direction of y-axis.

The line perpendicular to √3x – y + 5 = 0 is x + √3y + λ = 0  

It is given that the line x + √3y + λ = 0 cuts off an intercept of 4 units with the negative direction of the y-axis.

This means that the line passes through (0,-4).

So,

Let us substitute the values in the equation x + √3y + λ = 0, we get

0 – √3 (4) + λ = 0 
λ = 4√3

Now, substitute the value of λ back, we get

x + √3y + 4√3 = 0

∴ The required equation of line is x + √3y + 4√3 = 0.