Question

Find the points which trisect the line segment joining the points (0,0) and (9,12).​

Answer 1

$\huge{\bf{\underline{\underline{\green{Answer}}}}}$

See the attachment

The marked points are equal since P and Q are points which trisect it

So

P= $\frac{mx_{1}+nx_{2}}{m+n}$,$\frac{my_{1}+ny_{2}}{m+n}$

P= $\frac{1(0)+2(9)}{1+2}$,$\frac{1(0)+2(12)}{1+2}$

P= $\frac{18}{3}$,$\frac{24}{3}$

P=6,8

Similarly,

here m=2 and n=1

so

Q=3,4

Answer 2

Answer:

$\large\bold\red{(3,4)\:and\:(6,8)}$

Step-by-step explanation:

Given that,

• Line segment joining the points (0,0) and (9,12) is trisected by two points.

Now,

• Let the end points of line segment be A(0,0) and B(9,12).

Also,

• Let the points trisecting the line segment be C(x,y) and D(m,n).

Now,

we know that,

• Trisected means the line segment is divided into 3 parts of equal length.

Also,

We know that,

• the mid point of a line segment having it's ends (p,q) and (r,s) is given by, $\bold{(\frac{p+r}{2},\frac{q+s}{2})}$

Note:- Refer to the attachment for the figure.

Now,

We have,

C is the mid point of AD.

Therefore,

we get,

$= > \frac{m + 0}{2} = x \\ = > m = 2x \: \: \: .............(i)$

And

$= > \frac{n + 0}{2} = y \\ = > n = 2y \: \: \: \: \: .............(ii)$

Also,

D is the mid point of BC.

Therefore,

We get,

$= > \frac{x + 9}{2} = m \\ = > x = 2m - 9 \\ = > x = 2(2x) - 9 \: \: \: \: \: \: \: ...........from \: \:( i) \\ = > x = 4x - 9 \\ = > 4x - x = 9 \\ = > 3x = 9 \\ = > x = \frac{9}{3} \\ = > x = 3$

And

$= > \frac{y + 12}{2} = n \\ = > y = 2n - 12 \\ = > y = 2(2y) - 12 \: \: \: \: ...........from \: \: (ii) \\ = > y = 4y - 12 \\ = > 4y - y = 12 \\ = > 3y = 12 \\ = > y = \frac{12}{3} \\ = > y = 4$

Therefore,

$m = 2x = 2 \times 3 = 6$

And

$n = 2y = 2 \times 4 = 8$

Hence,

the required points are $\large\bold{(3,4)\:and\:(6,8)}$