Answer:
Step-by-step explanation:
Given
Required
Determine LN
Since LM is a bisector, then we have:
(See attachment for illustration)
Collect Like Terms
Solve for w
LN is calculated as thus:
Substitute 4 for w
Can someone please help me with this?!
1 answer:
To put the given line in standard form, a factor of 2 must be removed from each of the numbers.
Given line in standard form:
.. 3x -y = 9
The perpendicular line through point (h, k) can be found from the formula
.. (-given coefficient of y)*(x -h) +(given coefficient of x)*(y -k) = 0
.. 1*(x -(-7)) +3*(y -4) = 0 . . . . . . . . . (h, k) = (-7, 4)
Eliminating parentheses, you have
.. x +7 +3y -12 = 0
In standard form, the equation is
.. x +3y = 5
_____
The slopes of perpendicular lines are the negative reciprocal of each other. That is, their product is -1. For line ax +by = c, the slope is given by -a/b. The negative reciprocal of that is b/a, corresponding to line bx -ay = (some constant).
When the constant is zero, the line goes through the origin. You can translate the line to make it go through any point (h, k) by writing the equation as
.. b(x -h) -a(y -k) = 0
For the line to be in "standard form", the leading coefficient must be positive. If b < 0, then you want to write
.. -b(x -h) +a(y -k) = 0
as we have above.
Of course, this can be rearranged to
.. -bx +ay = -bh +ak . . . . standard form (when b<0)
Given line in standard form:
.. 3x -y = 9
The perpendicular line through point (h, k) can be found from the formula
.. (-given coefficient of y)*(x -h) +(given coefficient of x)*(y -k) = 0
.. 1*(x -(-7)) +3*(y -4) = 0 . . . . . . . . . (h, k) = (-7, 4)
Eliminating parentheses, you have
.. x +7 +3y -12 = 0
In standard form, the equation is
.. x +3y = 5
_____
The slopes of perpendicular lines are the negative reciprocal of each other. That is, their product is -1. For line ax +by = c, the slope is given by -a/b. The negative reciprocal of that is b/a, corresponding to line bx -ay = (some constant).
When the constant is zero, the line goes through the origin. You can translate the line to make it go through any point (h, k) by writing the equation as
.. b(x -h) -a(y -k) = 0
For the line to be in "standard form", the leading coefficient must be positive. If b < 0, then you want to write
.. -b(x -h) +a(y -k) = 0
as we have above.
Of course, this can be rearranged to
.. -bx +ay = -bh +ak . . . . standard form (when b<0)
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Graph each points:
f(x) = -13x + 1
when x = 0, y = 1
f(0) = -13(0) + 1
f(0) = 0 + 1
f(0) = 1
when x = 1, y = -14
f(1) = -13(1) + 1
f(1) = -13 + 1
f(1) = -14
when x = 2, y = -25
f(2) = -13(2) + 1
f(2) = -26 + 1
f(2) = -25
etc.
Graph each point and connect them.
