Question

write an equation of the line containing the specified point and perpendicular to the indicated line (-4,5) 7x-2y=1

Answer 1

Equation of a line given $7x-2y = 1$

We have to find an equation of a line which is perpendicular to the given line.

If the general equation of a line is $y=mx+c$, m is the slope of the line there. And the slope of the perpendicular line will be the negative reciprocal of m which is $(-\frac{1}{m})$.

So first here we have to make the given equation as $y=mx+c$

$7x-2y = 1$

First we have to move 7x to the right side by subtracting it from both sides.

$-7x+7x-2y =-7x+1$

$-2y = -7x+1$

Now to get y we have to move -2, by dividing it on both sides.

$\frac{(-2y)}{(-2)} = \frac{(-7x+1)}{(-2)}$

$y = \frac{(-7x)}{(-2)} + \frac{1}{(-2)}$

$y= (\frac{7}{2} )x - \frac{1}{2}$

So here the slope $m= \frac{7}{2}$

Now the slope for the perpendicular equation is

$-\frac{1}{m}= -\frac{1}{(7/2)}$

$-\frac{1}{m} = -\frac{2}{7}$

So slope of the perpendicular line is $-\frac{2}{7}$

We can write the perpendicular equation as

$y=(-\frac{2}{7} )x+c$

Now this equation is passing through the point (-4,5)

We have to plug in x = -4 and y = 5 in the line to get c.

$5= (-\frac{2}{7})(-4) + c$

$5= \frac{8}{7} +c$

$5-\frac{8}{7} = c$

$\frac{35}{7} -\frac{8}{7} =c$

$\frac{(35-8)}{7} = c$

$\frac{27}{7} =c$

So we have got the value of c. Now we can write the perpendicular equation as,

$y= (-\frac{2}{7})x+\frac{27}{7}$

$y = \frac{(-2x+27)}{7}$

$7y = -2x+7$

$7y+2x = -2x+2x+7$

$2x+7y = 7$

So we have got the required perpendicular line.

The equation of the perpendicular line is $2x+7y = 7$