Question

7y + 4x = 3; (-4, -7) Write the equation in slope intercept form of the line that is perpendicular to the graph of the equation and passes through the given point.

Answer 1

Solving the given equation for y results in  y = (-4/7) x + (a constant).

The new line has a slope that is the negative reciprocal of -4/7.  That slope is +7/4.

The new line passes thru (-4, -7).  Thus,

the equation for this new line is  y - (-7) = (7/4)(x - [-7]), or 

y + 7 = (7/4)(x+7)

Let's put this into slope-intercept form:

y = -7 + (7/4)x + (49/4), or        y = (7/4)x + (49/4) - 7,   or

y = (7/4)x + 21/4   (answer in slope-intercept form)

Answer 2

Answer:

$y=\frac{7}{4} x$

Step-by-step explanation:

we have the original line equation

$7y + 4x = 3$

clearing for y:

$7y=-4x+3\\y=\frac{-4}{7}x+ \frac{3}{7}$

Now we have an equation of the form slope- intercept:

$y=mx+b$

where m is the slope and b is the y-intercept.

thus, the slope of the original line is:

$m=\frac{-4}{7}$

Now to find the new line, since it has to be perpendicular their slopes must satisfy the following:

$m*m_{1}=-1$

where m is the slope of the original line, and m1 is the slope of the new line:

$\frac{-4}{7}*m_{1}=-1\\ m_{1}=\frac{-1*7}{-4}\\ m_{1}=\frac{7}{4}$

this is the slope of the new perpendicular line that passes trough the point (-4,-7), so now we use the point slope equation to find the equation of said line:

$y-y_{1}=m_{1}(x-x_{1})$

where we know $m_{1}=\frac{7}{4}$, and from the point (-4,-7) $x_{1}=-4, y_{1}=-7$

so we have:

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

and we clear for y to leave the equation in the slope intercept form:

$y=\frac{7}{4} (x+4)-7\\y=\frac{7}{4} x+\frac{7}{4}*4 -7\\\\y=\frac{7}{4} x+7-7\\y=\frac{7}{4} x$