Math, asked by mallikarjunaraomalle, 5 hours ago

find the equation of a line whose slope is -2/3 and passing through the line Segments passing through the points (1,2) (-3,5) in the ratio -4:3​

Answers

Answered by manishgupta8462
0

Answer:

2x-3y-5=0

Step-by-step explanation:A = (1, 2), B = (4, -3)

Point P divided seg AB internally in the ratio 3 : 4

Let, P = (x, y)

By section formula for internal division,

x = (mx2 + nx1)/ (m+n)

= 3(4) + 4(1)/ (3+4)

= (12 + 4)/ 7

= 16/7

y = (my2 + ny1)/ (m+n)

= 3(-3) + 4(2)/ (3+4)

= (-9 + 8)/7

= -1/7

P = (16/7, -1/7)

The line having slope 2/3 passes through the point P (16/7, -1/7).

The equation of the line by slope point form is,

(y - y1) = m(x-x1)

(y - (-1/7)) = 2/3((x- 16)/7)

y + 1/7 = 2/3 ((7x-16)/7)

(7y +1)/7 = 2/3((7x -16)/7)

7y +1 = 2/3 (7x-16)

7y +1 = (14x -32)/3

14x -32 = 21y +3

14x -21y -32 -3 = 0

14x -21y -35 = 0

7(2x -3y -5) = 0

2x -3y -5 = 0

The equation of the required line is 2x - 3y - 5 = 0

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