Question

Find the equation of the line which is perpendicular to 4𝑦=5𝑥−8 and passing through (2,3).

Answer 1

$$4𝑦=5𝑥−8=>y=\dfrac{5}{4}x-2$$

$$slope_1=m_1=\dfrac{5}{4}$$

If two lines with slopes $m_1$ and $m_2$ are perpendicular then $m_1m_2=-1.$

Then $\dfrac{5}{4}m_2=-1.$

$$slope_2=m_2=-\dfrac{4}{5}$$

The equation of the second line is

$$y=-\dfrac{4}{5}x+b$$

This line passes through the point $(2, 3)$

$$3=-\dfrac{4}{5}(2)+b=>b=y=\dfrac{23}{5}x$$

The equation of the line which is perpendicular to $4𝑦=5𝑥−8$ and passing through $(2,3)$ is

$$y=-\dfrac{4}{5}x+\dfrac{23}{5}$$

Or

$$5y=-4x+23$$