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प्रश्न
Find the point on y-axis whose distances from the points A (6, 7) and B (4, -3) are in the ratio 1: 2.
उत्तर १
Let the required point on y-axis be P (0, y).
PA = `sqrt((0 - 6)^2 + (y - 7)^2)`
= `sqrt(36 + y^2 + 49 - 14y)`
= `sqrt(y^2 - 14y + 85)`
PB = `sqrt((0 -4)^2 + (y + 3)^2)`
= `sqrt(16 + y^2 + 9 + 6y)`
= `sqrt(y^2 + 6y + 25)`
From the given information, we have:
`"PA"/"PB" = (1)/(2)`
`"PA"^2/"PB"^2 = (1)/(4)`
`(y^2 - 14y + 85)/(y^2 + 6y + 25) = (1)/(4)`
4y2 - 56y + 340
= y2 + 6y + 25
3y2 - 62y + 315
= 0
y = `(62 ± sqrt(3844 - 3780))/(6)`
y = `(62 ± 8)/(6)`
y = `9,(35)/(3)`
Thus, the required points on y-axis are (0, 9) and `(0,(35)/(3))`.
उत्तर २
Let the required point on y-axis be P (0, y).
PA = `sqrt((0 - 6)^2 + (y - 7)^2)`
= `sqrt(36 + y^2 + 49 - 14y)`
= `sqrt(y^2 - 14y + 85)`
PB = `sqrt((0 -4)^2 + (y + 3)^2)`
= `sqrt(16 + y^2 + 9 + 6y)`
= `sqrt(y^2 + 6y + 25)`
From the given information, we have:
`"PA"/"PB" = (1)/(2)`
`"PA"^2/"PB"^2 = (1)/(4)`
`(y^2 - 14y + 85)/(y^2 + 6y + 25) = (1)/(4)`
4y2 - 56y + 340 = y2 + 6y + 25
3y2 - 62y + 315 = 0
y = `(62 ± sqrt(3844 - 3780))/(6)`
y = `(62 ± 8)/(6)`
y = `9,(35)/(3)`
Thus, the required points on y-axis are (0, 9) and `(0,(35)/(3))`.
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