# IN WHAT RATIO DOES THE LINE x-y-2=0 divide the line segment joining (3,-1) and (8,9) ?

2/3

• 5

can u plz solve it..wid detail???

• 3

let t line x-y-2=0 divid t line segment in t ratio k : 1 at point C

so coordinates of C are

{ ( 8k +3 / k+1) , ( 9k-1 / k+1 ) }

C lies on x - y - 2 = o

so, subs coordinates of x,y in tis eq.

( 8k +3 / k+1) - ( 9k-1 / k+1 ) - 2 = 0

8k +3 - 9k +1 - 2k -2 / k+1 = 0

-3k +2 / k+1 = 0

-3k + 2 = 0

=> k = 2/3

i.e. 2 : 3

• 106

asif2998 who do you think you are!!!! dont you dare judge someone!!

had you been as smart as you THINK you are, you woudnt even be on this site !

• 10
thnkz
• -6
Consider the "journey" from (3, -1) to (8, 9).
The x-coordinate increases by 5, while the y-coordinate increases by 10.
Quick mental trial-and-error tells us that the two points lie on the line y = 2x - 7

Now we want the point of intersection of y = 2x - 7 and x - y - 2 = 0

Using substitution:
x - y - 2 = 0
x - (2x - 7) - 2 = 0
-x + 5 = 0
x = 5

y = 2(5) - 7
y = 3

The point of intersection is (5, 3).
The three points under consideration are (3, -1), (5, 3) and (8, 9).
Just dealing with x-coordinates: from 3 to 5 = 2 units, and from 5 to 8 = 3 units.

SOLUTION: the straight line x - y - 2 = 0 divides the line segment (3,-1) to (8,9) in the ratio 2:3
• 11
k = 2/3
k= 2:3
• 5
Aishwary answer 2: 3 is write
• -8
2/3
• 1
Ratio is 2:3
• -1
Ratio 2:3
• 0  