State the equation of the line that is perpendicular to 4x-3y=10 through the point (-2,4)

1 answer:

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The equation of the line that is perpendicular to 4x - 3y = 10 through the point (-2,4) is $y-4=\frac{-3}{4}(x+2)$

Solution:

Given, line equation is 4x – 3y = 10

We have to find a line that is perpendicular to 4x – 3y = 10 and passing through (-2, 4)

Now, let us find the slope of the given line,

$\text { Slope of a line }=\frac{-\mathrm{x} \text { coefficient }}{\mathrm{y} \text { coefficient }}=\frac{-4}{-3}=\frac{4}{3}$

We know that, slope of a line $\times$ slope of perpendicular line = -1

$\begin{array}{l}{\text { Then, } \frac{4}{3} \times \text { slope of perpendicular line }=-1} \\\\ {\rightarrow \text { slope of perpendicular line }=-1 \times \frac{3}{4}=-\frac{3}{4}}\end{array}$

Now, slope of our required line = $\frac{-3}{4}$ and it passes through (-2, 4)

The point slope form is given as:

$\begin{array}{l}{y-y_{1}=m\left(x-x_{1}\right) \text { where } m \text { is slope and }\left(x_{1}, y_{1}\right) \text { is point on the line. }} \\\\ {\text { Here in our problem, } m=-\frac{3}{4}, \text { and }\left(x_{1}, y_{1}\right)=(-2,4)} \\\\ {\text { Then, line equation } \rightarrow y-4=-\frac{3}{4}(x-(-2))}\end{array}$