find the equation of the straight line which has a gradient of 2 and which passes through the point (2,1).
1. find the equation of the straight line which has a gradient of 2 and which passes through the point (2,1).
y-y1=m(x-x1)
y-1=2(x-2)
y=2x-4+1
y=2x-3
mungkin gitu(?)
2. 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
3. Equation line with gradient 4 and passes through the point (3,6) is
Jawaban:
jawaban sudah tersedia pada lampiran
4. Determine the equation of straight line: passing through (3,6) and parallel to the line 2y + 2x = 3
Penjelasan dengan langkah-langkah:
langkah pengerjaan ada pada
5. 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
6. 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
7. Equation of Straight Line 1. Determine the equation of straight line passes point A (-3, 5) and having gradient of 3. I Dan JELASKAN caranya!
Jawab:
y = 3x + 14Penjelasan dengan langkah-langkah:
Let the gradient of a straight line taken by m.
Given that the straight line passes a specific point A(x₁, y₁), the equation can be found using this formula:
y – y₁ = m(x – x₁)
We have: m = 3, point A(–3, 5) => x₁ = –3, y₁ = 5.
Plugging those values into the formula, we can get that:
y – 5 = 3[x – (–3)]
y – 5 = 3(x + 3)
y – 5 = 3x + 9
y = 3x + 9 + 5
y = 3x + 14
Verify this equation by plugging the values of x₁ and y₁:
5 = 3(–3) + 14
5 = –9 + 14
5 = 5 (correct)
∴ Conclusion: The equation of mentioned straight line is:
y = 3x + 14
8. The linear equation of a line passes through two points at (-2,4) and (6,3) is.
Jawaban:
y = 1/8x + 15/4
Penjelasan:
I've attached an image of the working for reference.
Note that a linear equation means that it is in the form of ax + b
1. First, what you need to do is find the gradient of the line using the two coordinates provided. For me, I find it easier to remember when I label the coordinates (which are in the form of (x,y)) with y2, y1, x2, and x1. It doesn't matter which coordinate you choose as the first pair and which pair you choose as the second.
2. Afterwards, use the general equation of a straight line, which is y = mx + c, and substitute the m (gradient) in the equation with the gradient you previously calculated.
3. Next, substitute the y and x in the equation with one of the given coordinates. For example, in my working I substituted the y with 3 and the x with 6 in order to find c. However, it doesn't matter which coordinate you use. Even if you use (-2,4), you would get the same result.
4. Once you've found c, just substitute the c in the equation with the c you've calculated and you have your answer!
9. Help asap.find the equation of line is thatline passes the point A -3,67 and hasgradient 1/2
pgl
-
I think the line passes point A (-3 , 6) with gradient 1/2 !
here your answer :
y - y₁ = m (x - x₁)
y - 6 = 1/2 (x + 3)
2y - 12 = x + 3
x - 2y + 15 = 0
10. the equation of the line through the points (1,5) and (2,3) is?
Penjelasan dengan langkah-langkah:
itu jawabannya
please make the best ok
11. The equation for the line running through P (-5,2) and Q (3, -4) is
EQUATION of a STRAIGHT LINE
The equation for the line running through P (-5,2) and Q (3,-4) is [tex]y=-\frac{3}{4}x-\frac{7}{4}[/tex].
-------------------
EXPLANATION
Equations of a straight line are written in the form
[tex]y = mx + c[/tex] or [tex]y - mx = c[/tex]
In which [tex]m[/tex] is the gradient of the line and [tex]c[/tex] is the [tex]y[/tex]-intercept or where the graph crosses the y-axis.
The value of [tex]m[/tex] is determined with the formula
[tex]m = \frac{y_{1}-y_{2}}{x_{1}-x_{2}}[/tex]
To find the equation we use
[tex]y-y_{1} = m (x-x_{1})[/tex]
Or, we can bypass finding [tex]m[/tex] with this formula
[tex]\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}[/tex]
--------------
Given:
P is (-5,2)
[tex]x_{1}=-5\\y_{1}=-2[/tex]
Q is (3,-4)
[tex]x_{2}=3\\y_{2}=-4[/tex]
Find:
the equation for the line running through P and Q
Step by step solution:
1. Determine which formula you want to use to find [tex]m[/tex] and [tex]c[/tex]. I use [tex]\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}[/tex] because it is simpler IMHO.
[tex]\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}} \\\frac{y-2}{(-4)-2}=\frac{x-(-5)}{3-(-5)}\\\frac{y-2}{-6}=\frac{x+5}{8}\\[/tex]
2. Cross multiply the two fractions
[tex]8(y-2)=-6(x+5)\\8y-16=-6x-30\\[/tex]
3. Put the above equation into the [tex](n)y = mx + c[/tex] form
[tex]8y=-6x-30+16\\8y=-6x-14[/tex]
3. Now, we need to put it in the [tex]y = mx + c[/tex] form. Do this by dividing the equation with 8.
[tex]\frac{8y=-6x-14}{8}\\y=-\frac{6}{8}x-\frac{14}{8}\\[/tex]
4. Simplify the fractions and we get
[tex]y=-\frac{3}{4}x-\frac{7}{4}[/tex]
Thus, the equation for the line running through P (-5,2) and Q (3,-4) is [tex]y=-\frac{3}{4}x-\frac{7}{4}[/tex]
or
[tex]y=-0,75x-1,75[/tex].
--------------------------
Pelajari lebih lanjut tentang persamaan garis lurus:
https://brainly.co.id/tugas/13526808
https://brainly.co.id/tugas/25363184
https://brainly.co.id/tugas/12472361
--------------------------
Detail Jawaban
Kelas : 8 SMP
Mapel : Matematika
Bab : 3 - Persamaan Garis Lurus
Kode : 8.2.3.1
Kata kunci : straight line equation
12. The line equation that is parallel to the line 3x-5y+4 = 0 and through the point (-10,2) is
Jawaban:
*
gradien sejajar
m2 = m1
*
mencari gradien
a/b
a = angka didepan x
b = angka didepan y
*
contoh
by = ax + c
maka gradiennya
a/b
*
tentukan m1 dari persamaan 3x - 5y + 4 = 0
ubah kedalam persamaan by = ax + c
3x - 5y + 4 = 0
-5y = -3x - 4
maka
gradiennya = -3/-5 = 3/5
*
tentukan m2
ingat m2 = m1 maka m2 = 3/5
*
rumus persamaan garis dengan gradien dan 2 buah titik
(y - y1) = m(x - x1)
*
maka persamaan garis dengan gradien 3/5 dan titik (-10 , 2)
x = -10 , y = 2
(y - y1) = m(x - x1)
(y - 2) = 3/5 (x - (-10))
y - 2 = 3/5 (x + 10)
y - 2 = 3/5 x + 6
y = 3/5 x + 6 + 2
y = 3/5 x + 8
semoga bermanfaat dan terimakasih
13. make an equation of the line that has a gradient in the middle and through (3,-4)
The equation of the line that has a gradient in the middle and through (3 , -4) is y = [tex]\tt - \dfrac {4}{3} x[/tex]
MoreExplanation:Gradient is a straight line that shows how steep the straight line is. The gradient is usually symbolized as m.
Gradient formula :
• through one coordinate, m = [tex]\tt \dfrac {y}{x}[/tex]
• through two coordinates, m = [tex]\tt \dfrac {y_2 - y_1}{x_2 - x_1}[/tex]
Equationofastraightlinethathasagradientformula:
• through one coordinate, [tex]\tt y - y_1 = m\ (x - x_1)[/tex]
• through two coordinates, [tex]\tt \dfrac {y - y_1}{y_2 - y_1} = \dfrac {x - x_1}{x_2 - x_1}[/tex]
• through two parallel lines, [tex]\tt m_1 = m_2[/tex]
• through two perpendicular lines, [tex]\tt m_1 \times m_2 = -1[/tex]
Tobeknown:
a line has a gradient in the middle and through (3 , -4)
Tobeasked:
The equation of the line = . . . . ?
Tobeanswered:
Because the straight line is through one coordinate,
m = [tex]\tt \dfrac {y}{x}[/tex]
m = [tex]\tt - \dfrac {4}{3}[/tex]
Equation :
[tex]\tt y - y_1 = m(x - x_1)[/tex]
y - (-4) = [tex]\tt - \dfrac {4}{3}[/tex] (x - 3)
y + 4 = [tex]\tt - \dfrac {4}{3} x - \dfrac {4}{3} \times -3[/tex]
y + 4 = [tex]\tt - \dfrac {4}{3} x + 4[/tex]
y = [tex]\tt - \dfrac {4}{3} x + 4 - 4[/tex]
y = [tex]\tt - \dfrac {4}{3} x[/tex]
Conclusion:
The equation of the line that has a gradient in the middle and through (3 , -4) is y = [tex]\tt - \dfrac {4}{3} x[/tex]
LearnMore:how to find the equation in which the gradient is known and through one coordinate https://brainly.co.id/tugas/25571198how to find the equation in which the gradient is known and through two coordinates https://brainly.co.id/tugas/25559882how to find the equation in such two parallel lines and through one coordinate https://brainly.co.id/tugas/25544954Detail:Subject : Math
Class : VIII (National School) or IX (Integrated School)
Chapter : Straight Lines Equation
Code : 8.2.3.1
Key Words : gradient, straight lines, equation
14. Line p is parallel to line 2x+y-1=0 line p passes through point (2, -3), so the linearequation of line p is... *
Jawab:
Penjelasan dengan langkah-langkah:
Garis p sejajar dengan garis 2x + y-1 = 0 garis p melewati titik (2, -3), sehingga
persamaan garis p adalah ...
15. a line cuts x-axis at (-5,0) and y-axis at (0,7).find1)the equation of this line2)the equation of another line parallel to this line and passes through point (10,5)
(y - y1) / (y2 -y1) = (x - x1) / (x2 - x 1)
(y - 0) / (7 - 0) = (x - -5) / (0 - - 5)
y / 7 = (x +5) / (0 + 5)
y / 7 = (x + 5) / 5
5y = 7 (x + 5)
5y = 7x + 35
y = (7/5)x + 7. => m = 7/5
2) y - y1 = m (x - x1)
y - 5 = 7/5 ( x - 10
y - 5== (7/5)x - 7
y = (7/5)x - 2
16. 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.
17. given that the line 2x+3y=k passes through the point (2,-4), find k.
substusi x = 2 , y= - 4
k = 2(2) + 3(-4)
k = 4 - 12
k = - 8
18. 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
19. Line g passes through point (2, 2) and perpendicular to line h with equation y = 3x4. The linear equation of line g is... *
Line g passes through point (2,2) and perpendicular to line h with equation y = 3x
mh = 3 then mg = -1/3
The linear equation of line g is
y - y1 = mg (x - x1)
y - 2 = -1/3 (x - 2)
3(y - 2) = -1(x - 2)
3y - 6 = -x + 2
x + 3y = 2 + 6
x + 3y = 8
The linear equation of line g is x + 3y = 8
20. An equation of the line that passes through (-3, 4) and parallel with line y = 3x + 5 is... A. y = 3x + 15 B. y = 3x + 13 C. y = 3x + 5 D. y = 3x − 13
An equation of the line that passes through (-3, 4) and parallel with line y = 3x + 5 is... (B) y = 3x + 13
PENJELASANDiketahui:
(-3, 4)Parallel with y = 3x + 5Ditanyakan:
Persamaan garisnyaJawab:
(-3,4)
x1 = -3y1 = 4y = 3x + 5
m = 3
y - y1 = m(x - x1)
y - 4 = 3(x - (-3))
y - 4 = 3(x + 3)
y - 4 = 3x + 9
y = 3x + 9 + 4
y = 3x + 13 (B)
So, the answer is (B)y=3x+13
The answer is B.y=3x+13
PEMBAHASANDiketahui:
Melalui titik (-3, 4)Sejajar dengan garis y = 3x + 5Ditanyakan:
Persamaan garisnyaJawab:
(-3,4)
=> x1 = -3, y1 = 4
[tex] \\[/tex]
y = 3x + 5
m = 3
[tex]\\[/tex]
y - y1 = m(x - x1)
y - 4 = 3(x - (-3))
y - 4 = 3(x + 3)
y - 4 = 3x + 9
y = 3x + 9 + 4
y = 3x + 13 (B)
[tex]\\ \\[/tex]
[tex]\boxed{\tt Answer ~ by: \blue{Blueriver}}[/tex]
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