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PLEASE HELP!! URGENT!!!! UNIT TEST!!!!

what is the equation of a line that is perpendicular to -3x + 4y = 4 and passes through the point (4,0)

Respuesta :

kudzordzifrancis

ANSWER

[tex]y=-\frac{4}{3}x+\frac{16}{3}[/tex]


Or

[tex]3y+4x=16[/tex]


EXPLANATION


Let us find the gradient of the line:

[tex]-3x+4y=4[/tex] by rewriting it in the slope intercept form.


[tex]\Rightarrow 4y=3x+4[/tex]


We divide through by 4 now;


[tex]\Rightarrow y=\frac{3}{4}x+1[/tex]


This is now in the form;

[tex]y=mx+c[/tex]

where

[tex]m=\frac{3}{4}[/tex] is he slope.


This implies that the slope of the line that is perpendicular to this line will be the negative reciprocal of [tex]m=\frac{3}{4}[/tex] .


Thus the perpendicular line has slope,

[tex]m=\frac{-1}{\frac{3}{4}}= -\frac{4}{3}[/tex].


Let the perpendicular line have equation,


[tex]y=mx+c[/tex]

When we substitute the slope we have;


[tex]y=-\frac{4}{3}x+c[/tex]

We substitute the point. [tex](4,0)[/tex] to find c.


[tex]0=-\frac{4}{3}(4)+c[/tex]


[tex]0=-\frac{16}{3}+c[/tex]


[tex]\frac{16}{3}=c[/tex]

We substitute c to obtain;


[tex]y=-\frac{4}{3}x+\frac{16}{3}[/tex]


Or

[tex]3y+4x=16[/tex]

alinakincsem

Answer:

y = -4/3x + 16/3 or 4x + 3y = 16

Step-by-step explanation:

We know that the standard form of equation for a line is:

[tex]y = mx +c[/tex] where [tex]m[/tex] is the slope of the line and [tex]c[/tex] is the y-intercept.

So writing the given equation in the standard form:

[tex]y = \frac{3}{4} x+1[/tex]

According to this equation, the slops (m) is 3/4. So the slope of line perpendicular to this would be negative reciprocal of 3 i.e. -4/3.

Putting the value of the slope and the given point to find the y-intercept:

[tex]y=mx+c[/tex]

[tex]0= -\frac{4}{3} (4)+c[/tex]

[tex]c = \frac{16}{3}[/tex]

Therefore, the equation of line perpendicular to -3x + 4y = 4 which passes through the point (4,0) is [tex]y = \frac{-4}{3}x+ \frac{16}{3}[/tex] or 4x + 3y = 16.

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