Question

Write an equation of the line containing the given point and perpendicular to the given line:

Answer 1

Answer:

$y=\frac{5}{9}x+\frac{-83}{9}$

Step-by-step explanation:

Given equation of line:

$9x+5y=3$

To find the equation of line perpendicular to the line of the given equation and passes through point (4,-7).

Writing the given equation of line in standard form.

Subtracting both sides by $9x$

$9x-9x+5y=3-9x$

$5y=3-9x$  

Dividing both sides by 5.

$\frac{5y}{5}=\frac{3}{5}-\frac{9x}{5}$  

$y=\frac{3}{5}-\frac{9x}{5}$  

Rearranging the equation in standard form $y=mx+b$

$y=-\frac{9x}{5}+\frac{3}{5}$  

Applying slope relationship between perpendicular lines.

$m_1=-\frac{1}{m_2}$

where $m_1$ and $m_2$ are slopes of perpendicular lines.

For the given equation in the form $y=mx+b$ the slope $m_2$can be found by comparing $y=-\frac{9x}{5}+\frac{3}{5}$ with standard form.

∴ $m_2=-\frac{9}{5}$

Thus slope of line perpendicular to this line $m_1$ would be given as:

$m_1=-\frac{1}{-\frac{9}{5}}$

∴ $m_1=\frac{5}{9}$

The line passes through point (4,-7)

Using point slope form:

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

Where $(x_1,y_1)\rightarrow (4,-7)$ and $m=m_1=\frac{5}{9}$

So,

$y-(-7)=\frac{5}{9}(x-4)$

Using distribution.

$y+7=(\frac{5}{9}x)-(\frac{5}{9}\times 4)$

$y+7=\frac{5}{9}x-\frac{20}{9}$

Subtracting 7 to both sides.

$y+7-7=\frac{5}{9}x-\frac{20}{9}-7$

Taking LCD to subtract fractions

$y=\frac{5}{9}x-\frac{20}{9}-\frac{63}{9}$

$y=\frac{5}{9}x+\frac{(-20-63)}{9}$

$y=\frac{5}{9}x+\frac{-83}{9}$

$y=\frac{5}{9}x-\frac{83}{9}$

Thus, the equation of line in standard form is given by:

$y=\frac{5}{9}x-\frac{83}{9}$