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Nana76 [90]
3 years ago
15

Write the equation of the line that is perpendicular to the given line and that passes through the given point

Mathematics
1 answer:
Alinara [238K]3 years ago
8 0
Slope-intercepted form:
y=mx+b
In this equation; "m" is the slope of the line.

We have this equation:
-x+5y=14
It´s slope-intercepted form would be:
-x+5y=14
5y=x+14
y=1/5 x+14/5  (slope-intercepted form)

And the slope in this line is: m=1/5

A line perpendicular to this line will be the next slope:
m`=-1/m; 
m´=-1/(1/5)=-5

Point-slope form:we need a point (x₀,y₀) and the slope:
y-y₀=m(x-x₀)
We have a point (-5,-2) and the slope, m=-5
y+2=-5(x+5)
y+2=-5x-25
y=-5x-25-2
y=-5x-27  (slope-intercepted form)

Answer: the equation would be y=-5x-27

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DFG and JKL are complementary angles. M DFG = x+5 and m JKL=x-9. The measure of DFG is ___, and the measure of JKL is ___.
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the sum of complementary angles = 90°, hence

x + 5 + x - 9 = 90

2x - 4 = 90 ( add 4 to both sides )

2x = 94 ( divide both sides by 2 )

x = 47

Thus ∠DFG = 47 + 5 = 52° and ∠JKL = 47 - 9 = 38°


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