Yaadannoo boqonnaa fi gaaffii shaakalaa

Boqonnaa 7 Herrega kutaa 10: Jiometrii Coordinate

Wanti fuula kana irra jiru hundi boqonnaa 7 kitaaba Herrega kutaa 10 tiif: wanta boqonnaan haguugu, yaadannoo gabaabaa, gaaffii shaakalaa deebii isaanii waliin, fi viidiyoo barnootaa bakka jirutti.

5
Kaardii: 10
Gaaffii shaakalaa boqonnaa 7

Boqonnaan 7 maal haguuga

1.Distance between two points
2.Division of a line segment
3.Equation of a line
4.Slopes of parallel and perpendicular lines
5.Applications

Yaadannoo gabaabaa boqonnaa 7

Kaardiiwwan fuula Herrega kutaa 10 irratti boqonnaa kana jala galmaa'an. Tokko tokkoon bani.

Bal'ina Rog-sadii Sarara Qindeessaa IrrattiFoormulaa

A = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|

A = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|

$A$: Baldhina kof-sadootti $\text{square units}$
$x_1, y_1$: Qindeessitoota fiixee tokkoffaa $\text{units}$
$x_2, y_2$: Qindeessitoota fiixee lammaffaa $\text{units}$
$x_3, y_3$: Qindeessitoota fiixee sadaffaa $\text{units}$

Yoom akka itti fayyadamtu

Qindeessitoonni gubbee rog-sadii sadan yoo kennaman bal'ina rog-sadiichaa shallaguuf itti fayyadami.

Of eeggadhu

Dhuma irratti yeroo hunda gatii gonkaa (absolute value) fudhadhu. Bal'inni zeeroo yoo ta'e qabxiileen sadan sarara tokko irra jiru.

Grade 10 Mathematics, fuula 328 irraa eegalee irraa qophaa'e; osoo hin ba'in dura yeroo lama mirkanaa'e

Foormulaa DistanceFoormulaa

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

$d$: Fageenya qabxiiwwan lamaan gidduu jiru $\text{units}$
$x_1, y_1$: Qindeessitoota qabxii tokkoffaa $\text{units}$
$x_2, y_2$: Qindeessitoota qabxii lammaffaa $\text{units}$

Yoom akka itti fayyadamtu

Fageenya qajeelaa qabxiilee lamaan sarara Cartesian irratti argaman gidduu jiru shallaguuf itti fayyadami.

Of eeggadhu

Qindeessitoota qabxiiwwan garaagaraa walitti hin makin. Iskuweer gochuun dura x irraa x, akkasumas y irraa y hir'isi.

Grade 10 Mathematics, fuula 328 irraa eegalee irraa qophaa'e; osoo hin ba'in dura yeroo lama mirkanaa'e

Sararoota Parallel fi PerpendicularGabatee wabiiParallel lines · Perpendicular lines

Line Relationship	Slope Condition	Geometric Meaning
Parallel lines	$m_1 = m_2$	Kofoota jallinaa walqixa
Perpendicular lines	$m_1 \cdot m_2 = -1$	Kofa qajeelaan walqaxxaamuru

Yoom akka itti fayyadamtu

Sararoonni lama Parallel yookiin Perpendicular ta'uu isaanii mirkaneessuuf yookiin hima walqixxoo isaanii shallaguuf itti fayyadami.

Of eeggadhu

Ulaagaan baay'ata Perpendicular sararoota dalgee fi dhaabbataa hin taane qofaaf hojjeta. Sararri dhaabbataan sarara dalgeetiif yeroo hunda Perpendicular dha.

Grade 10 Mathematics, fuula 328 irraa eegalee irraa qophaa'e; osoo hin ba'in dura yeroo lama mirkanaa'e

Foormulaa SectionFoormulaa

(x_0, y_0) = \left(\frac{p x_2 + q x_1}{p + q}, \frac{p y_2 + q y_1}{p + q}\right)

(x_0, y_0) = \left(\frac{p x_2 + q x_1}{p + q}, \frac{p y_2 + q y_1}{p + q}\right)

$x_0, y_0$: Qindeessitoota qabxii qoodaa $\text{units}$
$p : q$: Reeshoo qoodama keessoo $\text{dimensionless}$
$x_1, y_1$: Qindeessitoota qabxii dhumaa tokkoffaa $\text{units}$
$x_2, y_2$: Qindeessitoota qabxii dhumaa lammaffaa $\text{units}$

Yoom akka itti fayyadamtu

Qabxii kutaa sararaa tokko reeshiyoo p:q kennameen keessaan qoodu shallaguuf itti fayyadami.

Of eeggadhu

Reeshiyoo isa duraa p qabxii lammaffaa wajjin, reeshiyoo isa lammaffaa q immoo qabxii isa duraa wajjin baay'isi.

Grade 10 Mathematics, fuula 328 irraa eegalee irraa qophaa'e; osoo hin ba'in dura yeroo lama mirkanaa'e

Foormulaa Slope fi InclinationFoormulaa

m = \frac{y_2 - y_1}{x_2 - x_1} = \tan\theta

m = \frac{y_2 - y_1}{x_2 - x_1} = \tan\theta

$m$: Sloopii sararaa $\text{dimensionless}$
$\theta$: Kofa dagalee $^{\circ}$
$x_1, y_1$: Qindeessitoota qabxii tokkoffaa $\text{units}$
$x_2, y_2$: Qindeessitoota qabxii lammaffaa $\text{units}$

Yoom akka itti fayyadamtu

Dallaa sarara qajeelaa qabxiilee lama irraa yookiin kofa inni sarara x pozatiivii wajjin uumu irraa shallaguuf itti fayyadami.

Of eeggadhu

Sararri qajeelli dhaabbataan x1 fi x2 walqixa ta'e Slope hin qabu, zeeroo miti. Sarara ciisaa qofatu Slope zeeroo qaba.

Grade 10 Mathematics, fuula 328 irraa eegalee irraa qophaa'e; osoo hin ba'in dura yeroo lama mirkanaa'e

Gaaffii shaakalaa boqonnaa 7

Gaaffii filannoo 10 boqonnaa kana irratti barreeffaman, tokkoon tokkoon isaanii deebii fi sababa qabu. Deebii banuu dura yaali. Gaaffiin kun Ingiliffaa tiin.

Gaaffii 1

Find the distance between the points C(2, 3) and D(6, 0).

A.5 units
B.7 units
C.25 units
D.1 unit

Deebii fi sababa agarsiisi

Deebii: A 5 units

Sababa

Use the distance formula d = √((x₂ − x₁)² + (y₂ − y₁)²).

Here x₂ − x₁ = 6 − 2 = 4 and y₂ − y₁ = 0 − 3 = −3.

d = √(4² + (−3)²) = √(16 + 9) = √25 = 5 units.

Gaaffii 2

Find the midpoint of the line segment joining the points (2, 5) and (8, 1).

A.(5, 3)
B.(3, 2)
C.(10, 6)
D.(6, 3)

Deebii fi sababa agarsiisi

Deebii: A (5, 3)

Sababa

The midpoint formula is ((x₁ + x₂)/2, (y₁ + y₂)/2).

x = (2 + 8)/2 = 5 and y = (5 + 1)/2 = 3. So the midpoint is (5, 3).

Gaaffii 3

Find the gradient (slope) of the line passing through the points (2, 3) and (6, 11).

A.2
B.1/2
C.−2
D.4

Deebii fi sababa agarsiisi

Deebii: A 2

Sababa

Gradient m = (y₂ − y₁)/(x₂ − x₁) = (11 − 3)/(6 − 2) = 8/4 = 2.

Gaaffii 4

A line passes through the points (5, −2) and (5, 8). What is its gradient?

A.0
B.1
C.No gradient, because the line is vertical
D.−1

Deebii fi sababa agarsiisi

Deebii: C No gradient, because the line is vertical

Sababa

Both points have x = 5, so the x-coordinates are equal. The line is vertical (parallel to the y-axis). A vertical line has no slope, so it has no gradient.

Gaaffii 5

A point R divides the line segment joining A(1, 2) and B(4, 5) internally in the ratio 1 : 2. Find the coordinates of R.

A.(2, 3)
B.(3, 4)
C.(1, 2)
D.(3, 2)

Deebii fi sababa agarsiisi

Deebii: A (2, 3)

Sababa

Use the section formula with p = 1 and q = 2, where P(x₁, y₁) = (1, 2) and Q(x₂, y₂) = (4, 5).

x = (p·x₂ + q·x₁)/(p + q) = (1·4 + 2·1)/3 = 6/3 = 2.

y = (p·y₂ + q·y₁)/(p + q) = (1·5 + 2·2)/3 = 9/3 = 3.

So R = (2, 3).

Gaaffii 6

A straight line passes through the points (0, 5) and (2, 9). Which equation describes this line?

A.y = 2x + 5
B.y = 5x + 2
C.y = 2x − 5
D.y = −2x + 5

Deebii fi sababa agarsiisi

Deebii: A y = 2x + 5

Sababa

First find the slope: m = (9 − 5)/(2 − 0) = 4/2 = 2.

Use the slope-intercept form y = mx + b. The point (0, 5) gives b = 5, because x = 0 makes y = b.

So the equation is y = 2x + 5.

Gaaffii 7

Find the equation of the line that passes through the point (2, 1) and is parallel to the line y = 3x − 4.

A.y = 3x − 5
B.y = −3x + 7
C.y = 3x + 1
D.y = x/3 − 5

Deebii fi sababa agarsiisi

Deebii: A y = 3x − 5

Sababa

Parallel lines have the same slope, so m = 3 (the slope of y = 3x − 4).

Use the point-slope form y − y₁ = m(x − x₁) with (x₁, y₁) = (2, 1):

y − 1 = 3(x − 2), so y − 1 = 3x − 6, and y = 3x − 5.

Gaaffii 8

A line has gradient 2/3. What is the gradient of a line that is perpendicular to it?

A.−3/2
B.3/2
C.−2/3
D.2/3

Deebii fi sababa agarsiisi

Deebii: A −3/2

Sababa

Two non-vertical lines are perpendicular when m₁ × m₂ = −1.

So m₂ = −1 ÷ (2/3) = −1 × (3/2) = −3/2.

Gaaffii 9

Find the area of the triangle with vertices (2, 3), (0, −1) and (3, −5).

A.10 square units
B.20 square units
C.5 square units
D.12 square units

Deebii fi sababa agarsiisi

Deebii: A 10 square units

Sababa

Use A = ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|.

Substitute (x₁, y₁) = (2, 3), (x₂, y₂) = (0, −1), (x₃, y₃) = (3, −5):

2(−1 + 5) + 0(−5 − 3) + 3(3 + 1) = 8 + 0 + 12 = 20.

A = ½ × 20 = 10 square units.

Gaaffii 10

In a city grid, each unit is 1 km. A bus stop is at point A(−2, 3) and a school is at point B(4, −5). How far is the school from the bus stop in a straight line?