Question

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

Answer 1

ANSWER

$y=-\frac{4}{3}x+\frac{16}{3}$

Or

$3y+4x=16$

EXPLANATION

Let us find the gradient of the line:

$-3x+4y=4$ by rewriting it in the slope intercept form.

$\Rightarrow 4y=3x+4$

We divide through by 4 now;

$\Rightarrow y=\frac{3}{4}x+1$

This is now in the form;

$y=mx+c$

where

$m=\frac{3}{4}$ is he slope.

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

Thus the perpendicular line has slope,

$m=\frac{-1}{\frac{3}{4}}= -\frac{4}{3}$.

Let the perpendicular line have equation,

$y=mx+c$

When we substitute the slope we have;

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

We substitute the point. $(4,0)$ to find c.

$0=-\frac{4}{3}(4)+c$

$0=-\frac{16}{3}+c$

$\frac{16}{3}=c$

We substitute c to obtain;

$y=-\frac{4}{3}x+\frac{16}{3}$

Or

$3y+4x=16$

Answer 2

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:

$y = mx +c$ where $m$ is the slope of the line and $c$ is the y-intercept.

So writing the given equation in the standard form:

$y = \frac{3}{4} x+1$

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:

$y=mx+c$

$0= -\frac{4}{3} (4)+c$

$c = \frac{16}{3}$

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