Find the equation of the straight line which is perpendicular to the straight line 3x+4y=12 such that perpendicular distance from the origin is equal to the perpendicular distance from (3,2) on the given line

Step-by-step explanation:

First get the slope of the given line;

Given the equation of the line 3x+4y=12

Rewrite in standard form y = mx+c where m is the slope;

3x+4y=12

4y = -3x + 12

y = -3x/4 + 12/4

y = -3x/4 + 3

Hence the slope of the given line is -3/4

Since the required line is perpendicular to this line, the slope of the required line will be M = -1/(-3/4) = 4/3

Get the equation:

Substitute M = 4/3 and the point (3,2) into the point slope equation of a line as shown;

y-y0 = m(x-x0)

y - 2 = 4/3(x-3)

Cross multiply

3(y-2) = 4(x-3)

3y - 6 = 4x - 12

3y - 4x = -12 + 6

3y-4x = -6

Hence the required equation of the line is 3y-4x = -6

the answer is somewhere there guys it really . there im just not gonna answer

step-by-step explanation:

but i'm guessing ur first option tho

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.