Find the equation of the line passing through the point (6,3) that is perpendicular to the line 4x−5y=−10. Step 1: Find the slope of the line 4x−5y=−10.What would the perpendicular slope be?Step 2: Use the slope to find the y-intercept of the perpendicular line.Step 3: Write the equation of the line that passes through the point (6,3) that is perpendicular to the line 4x−5y=−10
1. Find the equation of the line passing through the point (6,3) that is perpendicular to the line 4x−5y=−10. Step 1: Find the slope of the line 4x−5y=−10.What would the perpendicular slope be?Step 2: Use the slope to find the y-intercept of the perpendicular line.Step 3: Write the equation of the line that passes through the point (6,3) that is perpendicular to the line 4x−5y=−10
Find the equation of the line passing through the point (6,3) that is perpendicular to the line 4x - 5y = -10?
Solution :
4x - 5y = -10
- 5y = 4x - 10
Divide both sides by -5
y = (4/-5)x - (10/-5)
y = (-4/5)x + 2
From the slope intercept form we get
m = - 4/5
The slope of a perpendicular is a negative inverse of - 4/5
m = (5/4)
Slope intercept form for the perpendicular line is
y = (5/4) x +b
Solve for b using (6, 3)
3 = (5/4) (6) + b
3 = 30/4 + b
b = 3 - 30/4
b = (12 - 30)/4
b = -18/4
b = -9/2
So the perpendicular line is y = (5/4)x - 9/2
The equation of the line passing through the point (6,3) that is perpendicular to the line 4x-5y=-10 is y = (5/4)x - 9/2.
2. Find the equation of the straight line with slope 2 passing through (-3, 4) .
Jawaban:
semoga bisa membantu ya
Penjelasan dengan langkah-langkah:
[tex]y - y1 = m(x - x1) \\ y - 4 = 2(x - ( - 3)) \\ y - 4 = 2x + 6 \\ y = 2x + 6 + 4 \\ y = 2x + 10 \\ 2x - y + 10 = 0[/tex]
3. Find the slope and y-intercept of the line with equation y – 7 = 3(x – 4).
Answer:
slope =
−
3
4
y intercept =
7
4
Explanation:
3x + 4y = 7
4y = -3x +7
y =
−
3
4
x +
7
4
comparing with y=mx+c
we get,
m= slope =
−
3
4
c = y intercept =
7
4
3x +4y = 7
3
7
x +
4
7
y =
7
7
3
7
x +
4
7
y = 1
x
7
3
+
y
7
4
= 1
comparing with
x
a
+
y
b
= 1
we get,
x intercept =
7
3
y intercept =
7
4
[tex]y = 3x - 12 + 7 \\ slope = 3 \\ y \: intercept \: x = 0 \\ y = 3(0) - 5 \\ (0. - 5)[/tex]
4. A straight line passes through the points A(- 4, 2) and B(2, - 9) Find : (a) the gradient of the line (b) the equation of the line
Jawab:
Penjelasan dengan langkah-langkah:
a.
A(- 4, 2)---> x₁ =-4 and y₁ =2
B(2, - 9) ---> x₂ =2 and y₂ = -9
gradient = (y₂-y₁)/(x₂-x₁) = (-9-2)/(2-4) = -11/-2 = 11/2
b.
y -y₁ =m (x-x₁)
y -2 = 11/2 (x+4)
2y - 4= 11 (x+4)
2y - 4 = 11x + 44
11x - 2y + 48 = 0
5. the gradient of the line perpendicular to line a is ...
Jawaban:
Perpendicular Lines (HIGHER TIER)
Penjelasan:
Two lines are perpendicular if one is at right angles to another- in other words, if the two lines cross and the angle between the lines is 90 degrees. If two lines are perpendicular, then their gradients will multiply together to give -1
Semogamembantu(๑•ᴗ•๑)♡mohonmaafjikajawabansayakurangtepat
6. She wants to make a line. She needs a ruler. The shape of ruller is ...
Jawaban:
Rectangle
Penjelasan:
The shape of a ruler is a Rectangle
7. The slope of the line through the points (4, h) and (h+3, 7) is -1/4 specify the value of hNo need to use ways
Jawaban:
Given two values for the input, \displaystyle {x}_{1}x
1
and \displaystyle {x}_{2}x
2
, and two corresponding values for the output, \displaystyle {y}_{1}y
1
and \displaystyle {y}_{2}y
2
—which can be represented by a set of points, \displaystyle \left({x}_{1}\text{, }{y}_{1}\right)(x
1
, y
1
) and \displaystyle \left({x}_{2}\text{, }{y}_{2}\right)(x
2
, y
2
)—we can calculate the slope \displaystyle mm, as follows
\displaystyle m=\frac{\text{change in output (rise)}}{\text{change in input (run)}}=\frac{\Delta y}{\Delta x}=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}m=
change in input (run)
change in output (rise)
=
Δx
Δy
=
x
2
−x
1
y
2
−y
1
where \displaystyle \Delta yΔy is the vertical displacement and \displaystyle \Delta xΔx is the horizontal displacement. Note in function notation two corresponding values for the output \displaystyle {y}_{1}y
1
and \displaystyle {y}_{2}y
2
for the function \displaystyle ff, \displaystyle {y}_{1}=f\left({x}_{1}\right)y
1
=f(x
1
) and \displaystyle {y}_{2}=f\left({x}_{2}\right)y
2
=f(x
2
), so we could equivalently write
\displaystyle m=\frac{f\left({x}_{2}\right)-f\left({x}_{1}\right)}{{x}_{2}-{x}_{1}}m=
x
2
−x
1
f(x
2
)−f(x
1
)
The graph in Figure 5 indicates how the slope of the line between the points, \displaystyle \left({x}_{1,}{y}_{1}\right)(x
1,
y
1
)
and \displaystyle \left({x}_{2,}{y}_{2}\right)(x
2,
y
2
), is calculated. Recall that the slope measures steepness. The greater the absolute value of the slope, the steeper the line is.
8. find the equation of the line parallel to the line y=2x+4 passing through the point (6,2)
Penjelasan:
y - y1 = m2(x - x1) y - 2 = 2(x - 6)y - 2 = 2x - 12y = 2x - 12 + 2y = 2x - 109. " if you ever find yourself stuck in the middle of the sea, I'll sail the world to find you."what does the line mean?
Jawaban:
I will help you in times of trouble
Penjelasan:
This is one of the famous song from bruno mars with the name "Count on me"
The whole song itself it's talking about a person that have a really close friend and are willing to give his/her help for his/her friend whenever they are in trouble so their friend can count him/her on any situation.
This line itself said that if his friend ever find themself stuck and don't know where to go (which is implied by the stuck in the middle of the sea meaning there is no way to go beside swimming the sea which is really dangerous and impossible) he/she will help them to find the way out even if it means that "he/she need to sail the world to find them"
Indeed, it is a beautiful song.
10. find the equation of the straight line which passes through the point (-2,4) and is parallel to the line y=2x +5.
y = 2x + 5
m = 2
y - b = m(x - a)
y - 4 = 2(x -(-2))
y - 4 = 2(x + 2)
y - 4 = 2x + 4
y = 2x + 4 + 4
y = 2x + 8
11. find the equation of the line through the -5.6 point and perpendicular to the line 3x-2y + 2 = 0
3x-2y+2=0
y=mx
-2y=-3x..
y=3/2x
m=3/2
3/2.m2=-1
m2=-2/3
y-y1=m(x-x1)
y-6=-2/3(x+2)
----------------------x3
3y-18=-2(x+2)
3y-18=-2x-4
2x+3y=-4+18
2x+3y=14
12. Find the slope of the straight line passing through A(-1, 3) and B(49, 3)
[tex]m= \: \frac{y2 - y1}{x2 - x1} \\ \\ m = \frac{3 - 3}{49 - ( - 1))} \\ \\ m = \frac{0}{50} \\ \\ m = 0[/tex]
slope = 0
13. Find the equation of the line joining the point (3,7) to the point of intersection of the lines x+y=4 and x-y+3=0
Find the equation of the line joining the point (3,7) to the point of intersection of the lines x+y=4 and x-y+3=0
jawab
mencari perpotongan
x+y=4 ⇒x+y=4
x-y+3=0 ⇒ x-y=-3
dieliminasi 2y=7
y=3,5
x+y=4
3,5+y=4
y=4-3,5
y=0,5
titik perpotongan adalah (3,5;0,5)
mencari persammanan dari titik(3,7) dan (3,5;0,5)
[tex][tex] \frac{x-3}{3,5-3} = \frac{y-7}{0,5-7} [/tex]
[tex] \frac{x-3}{0,5} = \frac{y-7}{-6,5} [/tex] [/tex]
-6,5x+19,5=0,5y-3,5
-6,5x-0,5y=-3,5-19,5
-6,5x-0,5y=-23
6,5x+0,5y=23
jadi persamaan adalah 6,5x+0,5y=23
14. If you ever find yourself stuck in the middle of the sea I'll sail the world to find you.” What does the line mean ?
jika kau terombang - ambing di tengah lautan, kan kulayari seluruh lautan tuk temukanmu
15. Find an equation of the tangent line to the hyperbola y=3/x at the point (3, 1).
Penjelasan dengan langkah-langkah:
y = 3x-1
m = -3/x² then m = -1/3
y -1 = -1/3(x - 3)
3y -3 = -x + 3
3y + x = 6
16. Let y+2x=3 be the equation of the line which meets the curve 2xsquare -3xy=14. Find the coordinates of the points of intersection of the line and the curve.
the intersect points are (-7/8 , 19/4) and (2,-1)
17. Find the equation of the line passing through the point (2,-5) and parallel to the line 5x + 7y = 46.tolong dibantu secepatnya, dan yang benermakasii
question:
Tentukan persamaan garis yang melalui titik (2,-5) dan sejajar dengan garis 5x + 7y = 46.
answer:
y -y1 =m(x - x1)
y -(-5)=(-5/7)(x -2)
y +5=(-5(x-2))/7
7(y +5)=-5x +10
7y+45=-5x +10
45-10=-5x -7y
-5x -7y=35
18. CG is parallel line to EHFind :a. The value of ab. The measure of angle <DCG
Penjelasan dengan langkah-langkah:
8a-45=6a-13
8a-6a=32
2a=32
a=16
sudutDCG=8a-45=8(16)-45=83⁰
19. find the equation of the line that is perpendicular to the line y = 3x - 1 through the point 12,4
Jawab:
x + 3y - 24 = 0Penjelasan dengan langkah-langkah:
The gradient (m) of a line perpendicular to another line is
-1 / the gradient of another line
y = 3x - 1, therefore m₁ = 3
-1 / (m₁) =
-1 / (3) =
m₀ = -1/3
Point passed = (12, 4)
x₁ = 12y₁ = 4With this equation, find the equation of line:
y-y₁ = m₀(x-x₁)
y-4 = -1/3(x-12)
y-4 = -1/3)x + 4
y = -1/3)x + 4 + 4
y = -1/3)x + 8
3y = -x + 24
or
x + 3y - 24 = 0 ✅嘉誠
20. The point M is the mid-point of the line joining the points (3. 7) and (-1, 1). Find the equation of the line through M which is parallel to the line y/3 + y/2 = 1
Jawaban:
y = 4
Penjelasan dengan langkah-langkah:
titik tengah M { (-1+3)/2 , (1+7)/2} , M (1 , 4)
y/3 + y/2 = 1
y = 6/5
garis sejajar melalui M adalah y = 4
Tidak ada komentar:
Posting Komentar