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
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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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