Two points have coordinates A (2,3) and B (6,7). If C(7,t) lies on the perpendicular bisector

@ Samiul Islam

I'm not a teacher, neither a professional. When I was young, I was a very bad pupil at school.

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You can solve it much more easily if you know that all points on the perpendicular bisector of AB (and only they) are equidistant from A and B.
In your case, AC² = (7-2)² + (t-3)² = t² - 6t + 34 and BC² = (7-6)² + (t-7)² = t² - 14t + 50.
Equating, you get 8t = 16 and t=2. So C is (7, 2).

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First being a perpendicular bisector you want to find the midpoint.
The midpoint=
(2+6)/2, (3+7)/2
=4, 5

Second you want to find the slope, to find the slope of the perpendicular line
slope=
(7-3)/(6-2)
=1
Therefore the slope of the perpendicular line is the negative reciprical of 1= -1

So, going from midpoint 4,5 to point C(7,t), the x co-ordinate is increased by 3. The slope being -1 means that the y co-ordinate will decrease by 3. This arrives at point c being (7,2) and t=2.

Use the two-point form of the equation of a line: y-y₁ = ((y₂-y₁)/(x₂-x₁)) * (x-x₁).
Here, y₁ = 3; y₂ = 7; x₁ = 2; x₂ = 6. Substitute these:
y - 3 = ((7-3)/(6-2)) * (x-2), or
y - 3 = (4/4) * (x-2), or
y - 3 = x - 2, or
y = x + 1. This is the line through A and B; its slope m = 1.

Perpendicular lines have slopes that are negative reciprocals, as in m and -1/m, so the perpendicular line's slope is -1/1 = -1.

Use the midpoint formula to find the midpoint of AB, which is where the perpendicular bisector intersects AB:
M = ((x₁+x₂)/2, (y₁+y₂)/2)
= ((2+6)/2, (3+7)/2)
= (8/2, 10/2)
= (4,5).

Now use this, along with the perpendicular line's slope, in the point-slope form of the equation of a line: y-y₁ = m(x-x₁).