Question

Therefore, the coordinates of the points of trisection of the line segm Do THIS Find the trisectional points of linejoining (2.6) and (-4,8). Find the trisectional points of line joining (-3.-5) and (-6.-8)​

Answer 1

Step-by-step explanation:

$Given :-</p><p></p><p>A rhombus shaped field has green grass for 18 cows of cows to graze.If each side of the rhombus is 30m.and its longer diagonal is 48 m.</p><p></p><p>To find :-</p><p></p><p>How much area of grass field will each cow be getting.</p><p></p><p>Solution :-</p><p></p><p>Here we know the property such as ,Two diagonals are perpendicularly bisects each other.</p><p></p><p>Here the diagonals divides the rhombus in four right angled triangles.</p><p></p><p>Now ,</p><p></p><p>From phythagorous theorem we get that,</p><p></p><p>( \frac{x}{2} ) {}^{2} + ( \frac{48}{2} ) {}^{2} = {30}^{2}(2x)2+(248)2=302</p><p></p><p>By solving this and multiplying 4 on both of the sides we get that,</p><p></p><p>{x}^{2} + {48}^{2} = {60}^{2}x2+482=602</p><p></p><p>Here we can use the polynomial identity for this equation to get perfect answer.</p><p></p><p>{x}^{2} = (60 + 48) \times (60 - 48) = {x}^{2} = {36}^{2}x2=(60+48)×(60−48)=x2=362</p><p></p><p>By squaring on both the sides we get that,</p><p></p><p>x=36m.</p><p></p><p>Now ,</p><p></p><p>Here we should find the area</p><p></p><p>We get that,</p><p></p><p>area = 4 \times \frac{1}{2} \times 18 \times 24 = 864 {m}^{2}area=4×21×18×24=864m2</p><p></p><p>Then,</p><p></p><p>By dividing the area and number of crow to graze</p><p></p><p>We get that,</p><p></p><p>= \frac{864}{18} = 48 {m}^{2}=18864=48m2</p><p></p><p>Therefore,</p><p></p><p>The area of grass field will each cow be getting is 48m^2.</p><p></p><p>Hope it helps u mate .</p><p></p><p>Thank you .</p><p></p><p>$

$Given :-</p><p></p><p>A rhombus shaped field has green grass for 18 cows of cows to graze.If each side of the rhombus is 30m.and its longer diagonal is 48 m.</p><p></p><p>To find :-</p><p></p><p>How much area of grass field will each cow be getting.</p><p></p><p>Solution :-</p><p></p><p>Here we know the property such as ,Two diagonals are perpendicularly bisects each other.</p><p></p><p>Here the diagonals divides the rhombus in four right angled triangles.</p><p></p><p>Now ,</p><p></p><p>From phythagorous theorem we get that,</p><p></p><p>( \frac{x}{2} ) {}^{2} + ( \frac{48}{2} ) {}^{2} = {30}^{2}(2x)2+(248)2=302</p><p></p><p>By solving this and multiplying 4 on both of the sides we get that,</p><p></p><p>{x}^{2} + {48}^{2} = {60}^{2}x2+482=602</p><p></p><p>Here we can use the polynomial identity for this equation to get perfect answer.</p><p></p><p>{x}^{2} = (60 + 48) \times (60 - 48) = {x}^{2} = {36}^{2}x2=(60+48)×(60−48)=x2=362</p><p></p><p>By squaring on both the sides we get that,</p><p></p><p>x=36m.</p><p></p><p>Now ,</p><p></p><p>Here we should find the area</p><p></p><p>We get that,</p><p></p><p>area = 4 \times \frac{1}{2} \times 18 \times 24 = 864 {m}^{2}area=4×21×18×24=864m2</p><p></p><p>Then,</p><p></p><p>By dividing the area and number of crow to graze</p><p></p><p>We get that,</p><p></p><p>= \frac{864}{18} = 48 {m}^{2}=18864=48m2</p><p></p><p>Therefore,</p><p></p><p>The area of grass field will each cow be getting is 48m^2.</p><p></p><p>Hope it helps u mate .</p><p></p><p>Thank you .</p><p></p><p>$

Answer 2

Answer:

(5,−3) and (3,4).

Step-by-step explanation:

Let A(7,−2) and B(1,−5) be the given points and P(x,y) and Q(x′,y′) are the points of trisection.

Step 1: Find the coordinate of P

Point P divides AB internally in the ratio 1:2

(x,y)=[1+21(1)+2(7),1+21(−5)+1(−2)]

=(31+14,3−5−4)=(315,3−9)=(5,−3)

Step 2: Find the coordinate of Q.

Point Q is the mid-point PB.

(x′,y′)=((5+1)/2,(−3−5)/2)=(3,−4)

Therefore, the coordinates of the points of trisection are (5,−3) and (3,4).