Question

The linear equation y = mx + b contains the point (4, -2) and is perpendicular to the

Answer 1

A linear equation is represented as: $y = mx + b$.

The equation of the line is: $y =3x - 14$

First, we calculate the slope (m) of $6x + 18y = 11$

Make y the subject

$18y = -6x + 11$

Divide by 18

$y = -\frac{1}{3}x + \frac{11}{18}$

In $y = mx + b$

$m \to$ slope

So, the slope of $y = -\frac{1}{3}x + \frac{11}{18}$ is:

$m = -\frac 13$

Since, the line is perpendicular to $6x + 18y = 11$

The slope of the line is:

$m_1 = -\frac{1}{m}$

So, we have:

$m_1 = -\frac{1}{-1/3}$

$m_1 = 3$

The line passes through (4,-2).

So, the equation is:

$y =m(x - x_1) + y_1$

This gives:

$y =3(x - 4) -2$

$y =3x - 12 -2$

$y =3x - 14$

Hence, the linear equation is:

$y =3x - 14$

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