Question

Which equations represent the line that is perpendicular to the line 5x − 2y = −6 and passes through the point (5, −4)? Select three options. y = –Two-fifthsx – 2 2x + 5y = −10 2x − 5y = −10 y + 4 = –Two-fifths(x – 5) y – 4 = Five-halves(x + 5)

Answer 1

For this case we have that by definition, the equation of a line in the slope-intersection form is given by:

$y = mx + b$

Where:

m: It's the slope

b: It is the cut-off point with the y axis

On the other hand we have that if two lines are perpendicular, then the product of their slopes is -1. So:

$m_ {1} * m_ {2} = - 1$

The given line is:

$5x-2y = -6\\-2y = -6-5x\\2y = 5x + 6\\y = \frac {5} {2} x + \frac {6} {2}\\y = \frac {5} {2} x + 3$

So we have:

$m_ {1} = \frac {5} {2}$

We find $m_ {2}:$$m_ {2} = \frac {-1} {\frac {5} {2}}\\m = - \frac {2} {5}$

So, a line perpendicular to the one given is of the form:

$y = - \frac {2} {5} x + b$

We substitute the given point to find "b":

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

Finally we have:

$y = - \frac {2} {5} x-2$

In point-slope form we have:

$y - (- 4) = - \frac {2} {5} (x-5)\\y + 4 = - \frac {2} {5} (x-5)$

ANswer:

$y = - \frac {2} {5} x-2\\y + 4 = - \frac {2} {5} (x-5)$

Answer 2

The equations represent the line that is perpendicular to the line 5x − 2y = −6 and passes through the point (5, −4) is $y+4=-2/5(x-5)$.

For this case we have that by definition,

The equation of a line in the slope-intersection form.

What is the slope intercept form?

The slope intercept form of line is $y=mx+c$.....(1)

Where m:is the slope

y- y coordinate

x- x intercept

c: y intercept

If two lines are perpendicular, then the product of their slopes is -1.

So$m_1\times m_2=-1$

The given line is,

$5x-2y=-6\\-2y=-5x-6.... (isolate y)

\\y=\frac{-5}{-2}x-\frac{6}{-2} ....(divide both side by 2)\

y=\frac{5}{2}x+3....(1)$

Compare the a equation 1 with slope intercept form

$m_1=5/2$ and c=3

If two lines are perpendicular then slope is

$m_1\times m_2=-1$

Then the value of

$m_2$ is given by

$m_2=\frac{-1}{m_1}$.......(2)

Use the value of $m_1$ in a equation(2)

We get,

$m_2=\frac{1}{-\frac{5}{2} }

=-2/5$$m_2

=-2/5$

So, a line perpendicular to the one given is of the form.$y=-2/5x+c$

We substitute the given point and $m_2$ into equation (1)to find c.

$-4=-2/5(5)+c\\-4=-2+c\\-4+2=c\\c=-2$

Therefore, we get by the slope intercept form

$(y-(-4))

=-\frac{2}{5}(x-5)\\y+4

=-2/5(x-5)$

Therefore we get the equations represent the line that is perpendicular to the line 5x − 2y = −6 and passes through the point (5, −4) is,$y+4=-2/5(x-5)$

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