We have a line y = 1/3x -6
We want a line that is perpendicular to this line
Perpendicular lines have slopes that multiply to -1
1/3 * m = -1
3 * 1/3 *m = -1 * 3
m = -3
The slope of the perpendicular line is -3
y = mx+b where m is the slope and b is the y intercept
y = -3x+b
We have a point on the line ( 7 ,-23)
Substitute this point into the equation
-23 = -3(7)+b
-23 = -21+b
Add 21 to each side
-23+21 = -21+21+b
-2 = b
y = -3x-2
In slope intercept form, the line perpendicular passing through (7,-23) is
y = -3x-2
Answer:
Only the First one, A is the correct answer. The other options such B and C are incorrect.
Step-by-step explanation:
Answer:
y = 1/3 x + 7
Step-by-step explanation:
Given; the line is passing through, (-6,5) and the slope is 1/3
We can get its equation;
We would take, a point (x,y) and the given point (-6,5)
Therefore; Since slope = Δy/Δx
Then;
(5-y)/ (-6-x) = 1/3
3(5-y) = -6 -x
15 - 3y = -6 - x
we get;
3y = 6 + x + 15
3y = x + 21
Therefore;
<u>y = 1/3 x + 7</u>
OK, so for this equation, your goal is to get the d, and ONLY the d, on one side of the equation. So, to start out, you need to multiply the entire equation, meaning both sides, by 8 because we are trying to get rid of those pesky fractions.
8(1/8(3d-2)=1/4(d+5))
The equation then turns into this because the 8 and 4 cancelled out with the 8.
1(3d-2)=2(d+5)
Now, we need to distribute the left over numbers into the parenthesis.
3d-2=2d+10
And finally, we need to get the d's on one side, and the numbers on the other, so we subtract 2d from both sides and add the 2 to both sides. They then cancel out to make
d=12
Hope it helps! :)
