Question

What is the equation of the line that is perpendicular to the like y=-4x + 7 and passes through the point (8,2)?

Answer 1

Answer:

$y=\frac{1}{4}x$

Step-by-step explanation:

Given equation of line:

$y=-4x+7$

To find the equation of line perpendicular to the line of the given equation and passes through point (8,2).

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 $y=-4x+3$ with standard form.

∴ $m_2=-4$

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

$m_1=-\frac{1}{-4}$

∴ $m_1=\frac{1}{4}$

The line passes through point (8,2)

Using point slope form:

$y_-y_1=m(x_-x_1)$

Where $(x_1,y_1)\rightarrow (8,2)$ and $m=m_1=\frac{1}{4}$

So,

$y-2=\frac{1}{4}(x-8)$

Using distribution.

$y-2=(\frac{1}{4}x)-(\frac{1}{4}\times 8)$

$y-2=\frac{1}{4}x-2$

Adding 2 to both sides.

$y-2+2=\frac{1}{4}x-2+2$

$y=\frac{1}{4}x$

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

$y=\frac{1}{4}x$