Question

Please solve the question in the attachment!
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Answer 1

ANSWER:-

Any point (x, y) of perpendicular bisector will be equidistant from A and B.

$\mathtt{ \to \sqrt{(x - 4)^{2} + (y - 5) {}^{2} }}$

$\mathtt{ = \sqrt{(x -2) {}^{2} + (y - 3) {}^{2} }}$

Solving we get -12x - 4y + 28 = 0

3x + y - 7 = 0

Therefore:-

d) 3x + y - 7 = 0

Answer 2

Given :-

The equation of the perpendicular bisector of line segment joining the points A (4, 5) and B (- 2, 3) is

$\red{\large\underline{\sf{Solution-}}}$

Let assume that line l be the perpendicular bisector of the line segment joining the points A (4, 5) and B (- 2, 3).

We know, perpendicular bisector l bisects the line segment as well as perpendicular to it.

So, Let assume that C (a, b) be the midpoint of AB.

We know,

Midpoint Formula

Let P(x₁, y₁) and Q(x₂, y₂) be two points in the coordinate plane and R(x, y) be the mid-point of PQ. Then, the coordinates of R will be:

$\sf\implies R = \bigg(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\bigg)$

So, Coordinates of midpoint of C (a, b) of the line segment joining the points A (4, 5) and B (- 2, 3) is

$\rm :\longmapsto\:(a, \: b) = \bigg(\dfrac{4 - 2}{2}, \: \dfrac{5 + 3}{2} \bigg)$

$\rm :\longmapsto\:(a, \: b) = \bigg(\dfrac{2}{2}, \: \dfrac{8}{2} \bigg)$

$\rm :\longmapsto\:(a, \: b) = (1, \: 4)$

So, Coordinates of C is (1, 4).

Now,

Slope of line segment joining the points A (4, 5) and B (- 2, 3) is given by

$\rm\implies \:Slope \: of \: AB = \dfrac{3 - 5}{ - 2 - 4} = \dfrac{ - 2}{ - 6} = \dfrac{1}{3}$

We know,

Two lines having slope m and M are perpendicular iff Mm = - 1.

As line l is perpendicular to AB.

$\rm\implies \:Slope \: of \: l = - 3$

Now, Equation of line which passes through the point (c, d) and having slope m is given by

$\purple{\rm :\longmapsto\:\boxed{\tt{ y - d \: = \: m(x - c) \: }}}$

So, Equation of line 'l' having slope - 3 which passes through the point (1, 4) is

$\rm :\longmapsto\:y - 4 = - 3(x - 1)$

$\rm :\longmapsto\:y - 4 = - 3x + 3$

$\\ \bf\implies \:\boxed{\tt{ \: \: 3x + y - 7 = 0\: \: }}$

is the required equation of perpendicular bisector.

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Different forms of equations of a straight line

1. Equations of horizontal and vertical lines

Equation of line parallel to x - axis passes through the point (a, b) is x = a.

Equation of line parallel to x - axis passes through the point (a, b) is x = a.

2. Point-slope form equation of line

Equation of line passing through the point (a, b) having slope m is y - b = m(x - a)

3. Slope-intercept form equation of line

Equation of line which makes an intercept of c units on y axis and having slope m is y = mx + c.

4. Intercept Form of Line

Equation of line which makes an intercept of a and b units on x - axis and y - axis respectively is x/a + y/b = 1.

5. Normal form of Line

Equation of line which is at a distance of p units from the origin and perpendicular makes an angle β with the positive X-axis is x cosβ + y sinβ = p.