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Question
The point A(2, 7) lies on the perpendicular bisector of line segment joining the points P(6, 5) and Q(0, – 4).
Options
True
False
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Solution 1
This statement is False.
Explanation:
If A(2, 7) lies on perpendicular bisector of P(6, 5) and Q(0, – 4),
Then AP = AQ
∴ AP = `sqrt((6 - 2)^2 + (5 - 7)^2`
= `sqrt((4)^2 + (-2)^2`
= `sqrt(16 + 4)`
= `sqrt(20)`
And A = `sqrt((0 - 2)^2 + (-4 - 7)^2`
= `sqrt((-2)^2 + (-11)^2`
= `sqrt(4 + 121)`
= `sqrt(125)`
So, A does not lies on the perpendicular bisector of PQ.
Solution 2
This statement is False.
Explanation:
If the point A(2, 7) lies on the perpendicular bisector of the line segment, then the point A satisfy the equation of perpendicular bisector.
Now, we find the equation of perpendicular bisector.
For this, we find the slope of perpendicular bisector.
∴ Slope of perpendicular bisector = `(-1)/("Slope of line segment joining the points" (5, -3) "and" (0, -4))`
= `(-1)/((-4 - (-3))/(0 - 5))` ...`[∵ "Slope" = (y_2 - y_1)/(x_2 - x_1)]`
= 5
[Since, perpendicular bisector is perpendicular to the line segment, so its slopes have the condition, m1 · m2 = – 1]
Since, the perpendicular bisector passes through the mid-point of the line segment joining the points (5, – 3) and (0, – 4).
∴ Mid-point of PQ = `((5 + 0)/2, (-3 - 4)/2) = (5/2, (-7)/2)`
So, the equation of perpendicular bisector having slope `1/3` and passes through the mid-point `(5/2, (-7)/2)` is
`(y + 7/2) = 5(x - 5/2)` ...[∵ Equation of line is (y – y1) = m(x – x1)]
⇒ 2y + 7 = 10x – 25
⇒ 10x – 2y – 32 = 0
⇒ 10x – 2y = 32
⇒ 5x – y = 16 ...(i)
Now, check whether the point A(2, 7) lie on the equation (i) or not
5 × 2 – 7
= 10 – 7
= 3 ≠ 16
Hence, the point A(2, 7) does not lie on the perpendicular bisector of the line segment.
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