Question

18. Show that the line segment joining the points A(-5, 8) and B(10, – 4) is trisected by the co-
ordinate axes. Also, find the points of trisection of AB.
Hint : Let x-axis divide AB in the ratio k : 1 at the point Px, 0). Let y-axis divide AB in the
ratio p : 1 at the point QO, y).]​

Answer 1

Answer:

Given,

A line segment joining the points

A (-5, 8) and B (10, -4)  

We need to prove that this line segment is trisected by the coordinate axes.

and., we have to find the coordinate points of trisection.

So, (As per the hint given in Question)

Let, x-axis divides the line AB in the ratio k : 1 at the point P (x, 0).

So, By the concept of section formula

$\to( x,\:0) = \bigg(\dfrac{(k)(10) + (1)(-5)}{k+1},\:\dfrac{(k)(-4)+(1)(8)}{k+1}\bigg)$

$\to( x,\:0) = \bigg(\dfrac{10k-5}{k+1},\:\dfrac{-4k+8}{k+1}\bigg)$

$\to x = \dfrac{10k-5}{k+1}\:,\:0=\dfrac{-4k+8}{k+1}$

taking

$\to 0=\dfrac{-4k+8}{k+1}$

$\to -4k+8=0$

$\to \boxed{\:k=2\:}$

Now,

$\to x = \dfrac{10k-5}{k+1}$

$\to x = \dfrac{10(2)-5}{(2)+1}$

$\to x = \dfrac{20-5}{3}$

$\to \boxed{\;x=5\;}$

Therefore,

• x-axis will divide AB in the ratio k:1 = 2:1 and
• coordinates of division will be ( 5, 0)

Now,

Let, y-axis divide AB in the ratio p : 1 at the point (0, y).

Again, by concept of section formula

$\to( 0,\:y) = \bigg(\dfrac{(p)(10) + (1)(-5)}{p+1},\:\dfrac{(p)(-4)+(1)(8)}{p+1}\bigg)$

$\to( 0,\:y) = \bigg(\dfrac{10 p- 5}{p+1},\:\dfrac{-4p+8}{p+1}\bigg)$

taking,

$\to 0 = \dfrac{10p-5}{p+1}$

$\to \boxed{\;p=\dfrac{1}{2}\;}$

now,

$\to y = \dfrac{-4p+8}{p+1}$

$\to y = \dfrac{-4\times(\dfrac{1}{2})+8}{(\dfrac{1}{2})+1}$

$\to y = \dfrac{-2+8}{(\dfrac{3}{2})}$

$\to \boxed{\;y=4\;}$

Therefore,

• y-axis divide AB in the ratio p:1 = 1:2, and
• coordinates of division will be ( 0, 4)

Refer to the attachment for figure.

Answer 2

Let P and Q be the point of trisection of the line segment joining the points A (-5, 8) and B (10, −4).

So, AP = PQ = QB

We have AP : PB = 1 : 2 Co-ordinates of the point P are

$\sf : \implies \bigg( \frac{1 \times 10 + 2 \times ( - 5) }{1 + 2} , \frac{1 \times ( - 4) + 2 \times 8}{1 + 2} \bigg) \\ \\ \sf : \implies \bigg( \frac{10 - 10 }{3} , \frac{16 - 4}{3} \bigg) \\ \\ \sf : \implies \bigg( \frac{0}{3} , \cancel{\frac{12}{3} } \bigg) \\ \\ \sf : \implies \bigg( 0, 4 \bigg)$

So point p lines on the y - axis

we have AQ : BQ = 2 : 1

Co-ordinates of the point Q are

$\sf : \implies \bigg( \frac{2 \times 10 + 1 \times ( - 5) }{2 + 1} , \frac{2 \times ( - 4) + 2 \times 8}{2 + 1} \bigg) \\ \\ \sf : \implies \bigg( \frac{20 - 5 }{3} , \frac{ - 8 + 8}{3} \bigg) \\ \\ \sf : \implies \bigg( \cancel{ \frac{15}{3}} , \frac{ 0}{3} \bigg) \\ \\ \sf : \implies \bigg( 5, 0 \bigg)$