What is the slope of a line perpendicular to the line whose equation is 6x+3y=-63

Mathematics

2 answers:

viva [34]2 years ago

7 0

Answer:

The slope of a line perpendicular to the line whose equation is 6x+3y=-63 will be: 1/2

Step-by-step explanation:

We know that the slope intercept-form of the line equation is

$y=mx+b$

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

Given the equation

$6x+3y=-63$

simplifying the equation to write in the slope-intercept form

$y=-2x-21$

Thus, the slope = -2

As we know that the slope of the perpendicular line is basically the negative reciprocal of the slope of the line, so

The slope of the perpendicular line will be:

$\frac{-1}{-2}=\frac{1}{2}$

Therefore, the slope of a line perpendicular to the line whose equation is 6x+3y=-63 will be: 1/2

nadya68 [22]2 years ago

4 0

Answer:

y=x/2−63

Step-by-step explanation: 6x+3y=-63 3y=6x-63= y=2/x-63

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