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3 years ago

Equation of a line that goes through point (4,2) and perpendicular to -x+y=1

Mathematics

2 answers:

mojhsa [17]3 years ago

3 0

Answer:

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aleksklad [387]3 years ago

3 0

Answer:

y= -x + 6

Step-by-step explanation:

-x + y =1 ---> y = x+1

the line perpendicular to this line should have negative inverse slope

the slope is 1 so the negative inverse of 1 is -1

set up

y = -x +b

since it goes through (4,2)

then we have

2 = -4 +b

b=6

so the linear equation is y = -x+6

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Given: KL ║ NM , LM = 45, m∠M = 50° KN ⊥ NM , NL ⊥ LM Find: KN and KL

Mice21 [21]

Answer:

$KL=45\tan 50^{\circ}\sin 50^{\circ}\approx 41.08\\ \\KN=45\sin 50^{\circ}\approx 34.47$

Step-by-step explanation:

Given:

KL ║ NM ,

LM = 45

m∠M = 50°

KN ⊥ NM

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1. Consider triangle NLM. This is a right triangle, because NL ⊥ LM. In this triangle,

LM = 45

m∠M = 50°

So,

$\tan \angle M=\dfrac{\text{opposite leg}}{\text{adjacent leg}}=\dfrac{NL}{LM}=\dfrac{NL}{45}\\ \\NL=45\tan 50^{\circ}$

Also

$m\angle LNM=90^{\circ}-50^{\circ}=40^{\circ}$ (angles LNM and M are complementary).

2. Consider triangle NKL. This is a right triangle, because KN ⊥ NM . In this triangle,

$NL=45\tan 50^{\circ}$

$m\angle KLN=m\angle LNM=40^{\circ}$ (alternate interior angles)

$m\angle KNL=90^{\circ}-40^{\circ}=50^{\circ}$ (angles KNL and KLN are complementary).

So,

$\sin \angle KNL=\dfrac{\text{opposite leg}}{\text{hypotenuse}}=\dfrac{KL}{LN}=\dfrac{KL}{45\tan 50^{\circ}}\\ \\KL=45\tan 50^{\circ}\sin 50^{\circ}\approx 41.08$

and

$\cos \angle KNL=\dfrac{\text{adjacent leg}}{\text{hypotenuse}}=\dfrac{KN}{LN}=\dfrac{KN}{45\tan 50^{\circ}}\\ \\KN=45\tan 50^{\circ}\cos 50^{\circ}=45\sin 50^{\circ}\approx 34.47$

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PIT_PIT [208]

Answer:

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