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

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

Explanation:

we have the original line equation

`7y + 4x = 3`

clearing for y:

`7y=-4x+3\\y=(-4)/(7)x+ (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=(-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:

`(-4)/(7)*m_(1)=-1\\ m_(1)=(-1*7)/(-4)\\ m_(1)=(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)=(7)/(4)`, and from the point (-4,-7)
`x_(1)=-4, y_(1)=-7`

so we have:

`y-(-7)=(7)/(4) (x-(-4))\\y+7=(7)/(4) (x+4)`

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

`y=(7)/(4) (x+4)-7\\y=(7)/(4) x+(7)/(4)*4 -7\\\\y=(7)/(4) x+7-7\\y=(7)/(4) x`