Question

find the ratio in which the straight line 2x+3y-20=0 divides the join of the point 2,3 and 2,10​

Answer 1

Answer:

Hope it helpful I tried my best

Step-by-step explanation:

Let the ratio be k:1

A(8,-9) B(2,1)

The point of intersection be X(x,y)

so , \begin{gathered}x= \frac{8+2k}{k+1} \\ y= \frac{k-9}{k+1}\end{gathered}

x=

k+1

8+2k

y=

k+1

k−9

So the coordinates will also satisfy the eqn, as it lies on it.

So,

2x+3y-5=0

⇒\begin{gathered}2 X \frac{8+2k}{k+1} + 3 X \frac{k-9}{k+1} - 5=0 \\ \frac{16+4k+3k-27}{k+1}=5 \\ -11+7k=5k+5 \\ 2k=16 \\ k=8\end{gathered}

2X

k+1

8+2k

+3X

k+1

k−9

−5=0

k+1

16+4k+3k−27

=5

−11+7k=5k+5

2k=16

k=8

So the ratio=8:1

Co-ordiates are :

x=8+2k/k+1=24/9

y=k-9/k+1=-1/9

Answer 2

Step-by-step explanation:

(2,3) and (2,10) occur on the line x=2.

So, to know the ratio

the equation of the straight line will be set on x=2

2×2 + 3y - 20 = 0

3y = 20 - 4

y = 16/3

Now, formulating ratio is quite a easy process

Ratio = ( y - y1 ) : ( y2 - y) [ because x=2 constant ]

= ( 16/3 - 3 ) : ( 10 - 16/3 )

= ( (16 - 9)/3) : ( (30-16)/3)

= 7 : 14

= 1 : 2