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प्रश्न
Write down the equation of a line whose slope is 3/2 and which passes through point P, where P divides the line segment AB joining A(-2, 6) and B(3, -4) in the ratio 2 : 3.
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उत्तर
Suppose that P(x, y) divides the line joining the points A(x1, y1) and B(x2, y2) internally in the ratio m : n.
Then the co-ordinates of P are given by the formula,
`x=(mx_2+nx_1)/(m+n)" and "y=(my_2+ny_1)/(m+n)`
`rArrx=(2(3)+3(-2))/(2+3)" and "y=(2(-4)+3(6))/(2+3)`
`rArrx=0" and "y=(-8+18)/5`
`rArrx=0" and "y=10/5`
`rArrx=0" and "y=2`
Thus P(x, y) ≡ P(0, 2)
Now we need to find the equation of the line
whose slope is m = 3/2 and passing though the point P(x1, y1) ≡ P(0, 2)
is y - y1 = m(x - x1)
`rArry-2=3/2(x-0)`
`rArr2(y-2)=3(x-0)`
`rArr2y-4=3x`
`rArr3x-2y+4=0`
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Solution:
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A(8, 9) = (x1, y1), B(1, 2) = (x2, y2) and P(6, 7) = (x, y)
Using Section formula of internal division,
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∴ 7m + 7n = `square` + 9n
∴ 7m – `square` = 9n – `square`
∴ `square` = 2n
∴ `m/n = square`
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