what are the coordinates of Q?
1. what are the coordinates of Q?
Jawaban:
The x coordinate is 0 at point Q and the y coordinate is 3 at point Q
Penjelasan dengan langkah-langkah:
moga membantu
2. 8. What are the coordinates of the turning point of the parabola y = x² - 4x + 3? * A. (-2,-1) B. (-2,1) C. (2,-1) D. (2,1)
Jawaban:
C. (2 -1)
Penjelasan dengan langkah-langkah:
Cari x
x= -b/2a
x= -(-4)/2(1)
x= 4/2
x= 2
Masukkan:
y= 2²-4(2)+3
y= 4-8+3
y= -4+3
y= -1
(2, -1)
3. Point P has coordinates (k,-3). Point Q has coordinates (2,-3). The length of PQ is 6.5 units and k < 0 Find the value of k.
The value of k is –4.5.
ExplanationDistance of Two Points
It is given that:
Point P has coordinates (k, –3).Point Q has coordinates (2, –3).The length of PQ is 6.5 units.k < 0Question: Find the value of k.
SIMPLER METHOD
Looking at both ordinates, we can get that point P and point Q are collinear, on the line of y = –3. It means the distance between point P and point Q is determined by the difference of their abscissas. Considering that k < 0, we get:
[tex]\begin{aligned}d_{PQ}&=x_Q-x_P\\\Rightarrow 6.5&=2-k\\k&=2-6.5\\\therefore\ k&=\bf{-}4.5\end{aligned}[/tex]
DISTANCE FORMULA METHOD
To find the value of k, we can also use the distance of two points formula derived from the Pythagorean theorem.
[tex]\begin{aligned}&d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\end{aligned}[/tex]
Rewriting that formula based on known facts we have, we get:
[tex]\begin{aligned}d_{PQ}&=\sqrt{\left(x_Q-x_P\right)^2+\left(y_Q-y_P\right)^2}\\\Rightarrow\;{d_{PQ}}^2&=\left(x_Q-x_P\right)^2+\left(y_Q-y_P\right)^2\\\Rightarrow {(6.5)}^2&={(2-k)}^2+\left(-3-(-3)\right)^2\\&={(2-k)}^2+0^2\\\Rightarrow\quad\;6.5&={}\pm\sqrt{(2-k)^2}\\&={}\pm(2-k)\\&=\begin{cases}2-k\\k-2\end{cases}\\\Rightarrow\qquad k&=\begin{cases}2-6.5\\6.5+2\end{cases}\\\Rightarrow\qquad k&=\begin{cases}\bf{-}4.5\\\bf8.5\end{cases}\end{aligned}[/tex]
Eliminating positive value of k, we get:
[tex]\therefore\ k=\bf{-}4.5[/tex]
CONCLUSION
∴ The value of k is –4.5.
4. A (-2,1) and B (6,5) are the opposite ends of the diameter of a circle. Find the coordinates of its centre.
we can find the coordinate of the circle centre by using the mean formula for the x and y
point A (-2,1)
x1 = -2 and y1= 1
point B (6,5)
x2 = 6 and y2 = 5
circle centre coordinate
[tex]x = \frac{x1 + x2}{2} = \frac{ - 2 + 6}{2} = \frac{4}{2} = 2[/tex]
[tex]y = \frac{y1 + y2}{2} = \frac{1 + 5}{2} = \frac{6}{2} = 3[/tex]
so, the centre of the circle is (2,3)
5. The line x - 2y =-4 cuts the x-axis and the y-axis at P and Q respectively. M is a point on PQ such that it is equidistant from the coordinate axes.ꜰɪɴᴅ :(i) the coordinates of P and of Q.(ii) the coordinates of M,(iii) the area of APMO.tolong
x - 2y = -4
(i) cuts the x-axis at P → y = 0
x - 2(0) = -4
x = -4
Thus, the coordinate of P (-4,0) ✔️
cuts the x-axis at Q → x = 0
0 - 2y = -4
y = 4/2
y = 2
Thus, the coordinate of Q (0,2) ✔️
(ii) M is a point on PQ such that it is equidistant from the coordinate axes → x = -y
-y -2y = -4
-3y = -4
y = 4/3
y = 1 ⅓
x = -y
x = -1 ⅓
Thus, the coordinate of M (-1 ⅓, 1 ⅓) ✔️
(iii) area of ∆ PMO
= ½. base. height
= ½. 4. 1 ⅓
= 2 ⅔ sq.unit ✔️
_______________
6. the gradient formula if there are two coordinates is....
ex = (2,1) and (3,4)
so, m = 4-1/3-2
= 3/1 or 3
7. Given that the quadratic function /(x) = 2x² -12x + 15-a(x+m) +n Find(a) the value of m and of n, hence state a coordinates of minimum point.
first,
for sure a=2, directly, as leading coef, alright?
then:
find the axis of symmetry: x=-12/(-2*2)=3
lha ini ternyata harus sama dengan :
its vertex m = -3; in this form: a(x+m)²+n,
how to find n: easy!
calculate f(3), and we get f(3) = -3
lha yaitu n-nya, iya bener itu,
n=-3,
try asking your teacher, why m=-3 not just 3,
then why n immediately taken from f(3).
I give u bonus:
attached.
8. convert the following point coordinates into polar coordinates E(3,√3)
~ TrigonometrY
the cartesian coordinates (x , y) = (3 , √3) will be changed into polar coordinates (r , α) by following this form :
[tex]r = \sqrt{ {x}^{2} + {y}^{2} } \\ r = \sqrt{ {3}^{2} + {( \sqrt{3} )}^{2} } \\ r = \sqrt{9 + 3} \\ r = 2 \sqrt{3} \\ \\ \alpha = arc \: \tan( \frac{y}{x} ) \\ \alpha = arc \: \tan( \frac{ \sqrt{3} }{3} ) \\ \alpha = 30°[/tex]
Hence , the polar coordinates of (3 , √3) is (2√3 , 30°)
9. The coordinates of 3 points are A(1, 1),B(-1, 4) and C(6, k). Find the value of k ifAB is perpendicular to BC.
Jawab:
[tex]gradientAB=-1:gradientBC[/tex]
[tex]\frac{yB-yA}{xB-xA} =-1:\frac{yC-yB}{xC-xB}[/tex]
[tex]\frac{yB-yA}{xB-xA} =-\frac{xC-xB}{yC-yB}[/tex]
[tex]\frac{4-1}{-1-1} =-\frac{6-(-1)}{k-4}[/tex]
[tex]\frac{3}{-2} =-\frac{7}{k-4}[/tex]
[tex]3(k-4)=-2(-7)[/tex]
[tex]3k-12=14[/tex]
[tex]k=8\frac{2}{3}[/tex]
Penjelasan dengan langkah-langkah:
10. The coordinates of X- intercept and Y-intercept of line 2y + 3x - 6 = 0 respectively are...
2y + 3x - 6 = 0 kalau salah bilang sama saya
11. If the coordinates A(-3,1),B(2,-4),C(7,1) is a corner points of square of abcd then the coordinates of D is
D(5,6),but im not sure
12. the line[tex]y = 18 - 3x[/tex]intersects the curve [tex] {y}^{2} = 9x[/tex]at points a and b.find1) the coordinates a and b2)the coordinates of the mid point of the line joining a and b3)the distance between a and b
(18-3x)²= 9x
324 - 108x +9x²=9x
9x²- 117x + 324 =0
x² - 13x + 36 = 0
(x - 9)(x - 4) = 0
x = 9 ; x = 4
jika x=9 maka y = 18-3(9) = -9
jika x=4 maka y = 18-3(4) =6
koordinat titik A(9, -9)
koordinat titik B(4, 6)
mid point (13/2, -3/2)
jarak titik A dan B √178
maaf jika salah.
:d
13. · The coordinates of the parallelogram ABCD are: A = (5,2), B = (2, 4), C = (6, 7) and D= (9,5). What is the length of the shorter diagonal of parallelogram ABCD?
Jawab:
[tex]\sqrt{26}[/tex]
Penjelasan dengan langkah-langkah:
Rumus jarak titik P(a, b) dengan titik Q(c, d) adalah [tex]|PQ|=\sqrt{(c-a)^2+(d-b)^2[/tex]
diagonal dari jajar genjang tersebut adalah AC dan BD. dengan rumus jarak dua titik didapat
[tex]|AC|=\sqrt{(6-5)^2+(7-2)^2}=\sqrt{1+25}=\sqrt{26}\\|BD|=\sqrt{(9-2)^2+(5-4)^2}=\sqrt{49+1}=\sqrt{50}\\\\[/tex]
karena yang diminta panjang diagonal terpendek, maka panjangnya adalah [tex]\sqrt{26}[/tex]
14. The coordinates of three points are A(1, 1), B(-1, 4) and C(h, k). Given that the gradients of AB, AC and BC are -3a, 3a and a respectively, find the values of h, k and a.
h = 5, k = 7, a = ½
Explanation/Penjelasan
The coordinates of three points are A(1, 1), B(–1, 4) and C(h, k).
Given that the gradients of AB, AC and BC are –3a, 3a and a respectively, we can get
[tex]\begin{aligned}&&m_{\rm AB}&=\frac{y_{\rm B}-y_{\rm A}}{x_{\rm B}-x_{\rm A}}\\\\&\Leftrightarrow&-3a&=\frac{4-1}{-1-1}\\&&&=\frac{3}{-2}\ =\ -3\cdot\frac{1}{2}\\\\&\therefore&a&=\bf\frac{1}{2}\\\\&\therefore&m_{\rm AC}&=3a={\frac{3}{2}}=\frac{y_{\rm C}-y_{\rm A}}{x_{\rm C}-x_{\rm A}}\\&\Leftrightarrow&\frac{3}{2}&=\frac{k-1}{h-1}\\&\Leftrightarrow&\!\!3h-3&=2k-2\quad...(i)\end{aligned}[/tex]
[tex]\begin{aligned}&&m_{\rm BC}&=a={\frac{1}{2}}=\frac{y_{\rm C}-y_{\rm B}}{x_{\rm C}-x_{\rm B}}\\&\Leftrightarrow&\frac{1}{2}&=\frac{k-4}{h-(-1)}\ =\ \frac{k-4}{h+1}\\&\Leftrightarrow&h+1&=2k-8\\&\Leftrightarrow&h&=2k-9\quad...(ii)\end{aligned}[/tex]
[tex]\begin{aligned}&&(ii)\to(i):\\&&\!\!\!\!3(2k-9)-3&=2k-2\\&\Leftrightarrow&6k-27-3&=2k-2\\&\Leftrightarrow&6k-30&=2k-2\\&\Leftrightarrow&6k-2k&=-2+30\\&\Leftrightarrow&4k&=28\\&&\therefore\ k&=\bf7\\\\&&(ii)\implies h&=2(7)-9\\&&&=14-9\\&&\therefore\ h&=\bf5\end{aligned}[/tex]
CONCLUSION/KESIMPULAN
∴ h = 5, k = 7, a = ½
gradient formula :
m = y2 - y1 / x2 - x1
15. A curve has an equation (2-akar x) pangkat 4 The normal at the point P(1, 1) and the normal at the point QC 9, 1) intersect at the point R. a) Find the coordinates of R b) Find the area of the triangle PQR.
Jawab:
a. R(7,13)
b. L=46 satuan luas
Penjelasan dengan langkah-langkah:
[tex]f(x)=(2-\sqrt{x})^4[/tex]
Garis normal adalah garis yang tegak lurus dengan garis singgung terhadap kurva. Jadi pertama-tama kita harus mencari garis singgung kurva pada titik P(1,1) dan Q(9,1). Untuk mencari
[tex]f(x)=(2-\sqrt{x})^4\\\\f'(x)=4(2-\sqrt{x})^3(-\frac{1}{2\sqrt{x}})\\f'(x)=-\frac{(2-\sqrt{x})^3}{2\sqrt{x}}[/tex]
Masukkan nilai P pada f'(x)[tex]P(x) :\\m=-\frac{(2-\sqrt{1})^3}{2\sqrt{1}}\\m=-\frac{1}{2}\\m_\perp=2[/tex]
Setelah ditemukan gradien tegak lurus, gunakan rumus [tex]\frac{y-y_1}{x-x_1}=m[/tex]
[tex]\frac{y-1}{x-1}=2\\y-1=2x-2\\y=2x-1[/tex]
Diperoleh [tex]P(x)=2x-1[/tex]
Masukkan nilai Q pada f'(x)[tex]Q(x):\\m=-\frac{(2-\sqrt{9})^3}{2\sqrt{9}}\\m=-\frac{(2-3)^3}{2\cdot3}\\m=-\frac{(-1)}{6}\\m=\frac{1}{6}\\m_\perp=-6[/tex]
Seperti cara pada P,
[tex]\frac{y-1}{x-9}=-6\\y-1=-6x+54\\y=-6x+55[/tex]
Diperoleh [tex]Q(x)=-6x+55[/tex]
Titik R adalah titik potong dari P(x) dan Q(x), sehingga
[tex]P(x)=2x-1\\Q(x)=-6x+55\\\\P(x)=Q(x):\\2x-1=-6x+55\\8x=56\\x=7[/tex]
Ditemukan bahwa nilai x dari R adalah 7, untuk menemuakn titik y nya maka masukkan x=7 ke dalam P(x) atau Q(x), hasilnya sama saja.
[tex]P(7)=2(7)-1\\P(7)=13\\y=13[/tex]
a) Koordinat titik R adalah (7, 13)
b) Luas segitiga
Kita bisa mencari luas segitiga hanya dengan koordinatnya menggunakan rumus
[tex]L=\frac{1}{2}\left[\begin{array}{ccc}1&1&1\\x_1&x_2&x_3\\y_1&y_2&y_3\end{array}\right][/tex]
[tex]L=\frac{1}{2}\left[\begin{array}{ccc}1&1&1\\1&9&7\\1&1&13\end{array}\right][/tex]
Cari determinan dari matriks
[tex]L=\frac{1}{2}((1\cdot9\cdot13+1\cdot7\cdot1+1\cdot1\cdot1)-(1\cdot7\cdot1+1\cdot1\cdot13+1\cdot9\cdot1))\\L=\frac{1}{2}(125-33)\\L=\frac{1}{2}(92)\\L=46[/tex]
Diperoleh luasnya sebesar 46 satuan luas
16. The curve y = 2x³ + ax + bx + 7 has a stationary point at the point (2, -13). a) Find the value of a and the value of b. b) Find the coordinates of the second stationary point on the curve. C)Determine the nature of the two stationary points. d) Find the coordinates of the point on the curve where the gradient is minimum and state the value of the minimum gradient. *If possible pls send it together with the working way, thank you so much
Jawab:
y = 2x³+ax²+bx+7 ⇒ substitute with (2, -13)
-13 = 2(2)³+a(2)²+b(2)+7 (I assume it's supposed ax², not ax)
-13 = 16+4a+2b+7
4a+2b=-36
b = -18-2a ⇒ [1]
a.
stationary point ⇒ gradient = 0 (y' = 0)
y = 2x³+ax²+bx+7
y' = 6x² + 2ax +b
0 = 6x² + 2ax +b ⇒ x = 2
0 = 6(2)² + 2a(2) +b ⇒ substitute with [1]
0 = 24 + 4a + (-18-2a)
2a = -6
a = -3
b = -18-2a = -18-2(-3) = -12
b.
y' = 6x² + 2ax +b
0 = 6x² + 2(-3)x +(-12)
0 = 6x² - 6x -12
0 = x² - x - 2
(x-2)(x+1) = 0
x₁ = 2
x₂ = -1
y = 2x³-3x²-12x+7
y(-1) = 2(-1)³-3(-1)²-12(-1)+7
= 14
coordinate (-1, 14)
c. (-1, 14) maximum point
(2, -13) minimum point
d.
y' = gradient
gradient min = y' min
= 6x² - 6x -12 min ⇒ y'' = 0
y'' = 12x - 6
0 = 12x - 6
x = 1/2
y(1/2) = 2(1/2)³-3(1/2)²-12(1/2)+7
= 1/2
coordinate ([tex]\frac{1}{2}[/tex] , [tex]\frac{1}{2}[/tex])
17. The un-shaded region of the following figure is the feasible region of a system of linear inequalities. The maximum value of (2x + 2y – 5) is attained at what point in the feasible region? a. Show the system of linear inequalities used in finding the maximum value b. Show the coordinates and the solutions made to get the coordinates at the corner points in the feasible region. c. Show the table of values 2x + 2y – 5.
Jawab:
chk it, rapikan, feedback
Penjelasan dengan langkah-langkah:
18. the coordinates of point G are (4,-2) and the gradient of the straight line GH is -2.The coordinate of point H could beA.(-1,6)B.(-1,8)C.(-1,10)D.(-1,12)
G(4 , -2)
m = -2
m = (yH - yG)/(xH - xG)
-2 = (yH - (-2))/(xH - 4)
-2 = (yH + 2)/(xH - 4)
-2(xH - 4) = yH + 2
8 - 2xH = yH + 2
2xH + yH = 8 - 2
2xH + yH = 6
xH = -1
2xH + yH = 6
2(-1) + yH = 6
-2 + yH = 6
yH = 6 + 2
yH = 8
The coordinate of point H could be
(xH , yH) ---> (-1 , 8)
19. Find the coordinates of the point A' which is the symmetric point of A(3,2) with respect to the line 2x + y - 12 =0 tolong donggg
2x+y-12=0
-y=2x-12
m¹=-2. m¹xm²=-1
-2xm²=-1
m²=-1/-2
m²=1/2
y-y¹=m(x-x¹)
y-2=1/2(x-3)
y=1/2x-3/2+2
2y-x=3.5
20. The coordinates of the point of intersection between y = x2 – 4x + 3 and y = 4x – 9 are …. A (4, -2) B (2, 1) C (6, 15) D (- 6,5) E (-2, 1)
Penjelasan dengan langkah-langkah:
x²-4x+3=4x-9
x²-8x+12=0
(x -6)(x-2)=0
x=6 V x=2
x y (x, y)
2 -1 (2, -1)
6 15 (6,15)
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