Question

The point A(2, 7) lies on the perpendicular bisector of line segment joining the points P(6, 5) and Q(0, – 4).

विकल्प

• True

• False

Answer 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.

Answer 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.