Question

Find the equation of the line that has $$x$$-intercept – 3 and is perpendicular to the line $$3x + 5y = 1$$. 

[Hint: The required line passes through $$(-3, 0)$$.]

Answer 1

The given line is $$3x + 5y = 1$$ 

$$5y = -3x + 1$$ 

$$y = -\frac{3}{5}x + \frac{1}{5}$$ 

Slope $$m_1 = -\frac{3}{5}$$

The slope of the required perpendicular line is:

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

$$= -\frac{1}{-\frac{3}{5}}$$

$$= \frac{5}{3}$$

Since the $$x$$-intercept is $$-3$$, the line passes through $$(-3, 0)$$. 

Equation of the line:

$$y - 0 = \frac{5}{3}(x - (-3))$$

$$3y = 5(x + 3)$$ 

$$3y = 5x + 15$$ 

$$5x - 3y + 15 = 0$$