Question

Find the equation of perpendicular bisector of the line segment joining the points (3, 4) and (-3, -2)​

Answer 1

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

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

Let assume that (3, 4) and (-3, -2) are represented as A and B

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

So, Let assume that l bisects the line segment joining the points A and B at C and l is perpendicular to AB.

Let assume that coordinates of C be (x, y).

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, using this

$\sf\implies (x,y) = \bigg(\dfrac{3 - 3}{2}, \dfrac{4 - 2}{2}\bigg)$

$\sf\implies (x,y) = \bigg(\dfrac{0}{2}, \dfrac{2}{2}\bigg)$

$\sf\implies (x,y) = (0, \: 1)$

Hence, Coordinates of C be ,(0, 1).

Now, we know that the slope of line joining the points (a, b) and (c, d) is given by

$\rm\implies \:\boxed{\tt{ Slope \: of \: line \: = \: \frac{d - b}{c - a} \: }}$

So, using this the slope of line joining A(3, 4) and B(-3, -2) is given by

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

We know,

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

Thus,

$\bf\implies \:Slope \: of \: line \: l \: = \: - \: 1$

Now, We know that,

Equation of line which passes through the point (a, b) and having slope m is given by y - b = m(x - a).

So, Equation of perpendicular bisector l which passes through the point (0, 1) and having slope - 1, is given by

$\rm :\longmapsto\:y - 1 = - 1(x - 0)$

$\rm :\longmapsto\:y - 1 = - x$

$\bf :\longmapsto\:x + y - 1 = 0$

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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.

Answer 2

Step-by-step explanation:

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

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

Let assume that (3, 4) and (-3, -2) are represented as A and B

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

So, Let assume that l bisects the line segment joining the points A and B at C and l is perpendicular to AB.

Let assume that coordinates of C be (x, y).

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, using this

$\sf\implies (x,y) = \bigg(\dfrac{3 - 3}{2}, \dfrac{4 - 2}{2}\bigg)$

$\sf\implies (x,y) = \bigg(\dfrac{0}{2}, \dfrac{2}{2}\bigg)$

$\sf\implies (x,y) = (0, \: 1)$

Hence, Coordinates of C be ,(0, 1).

Now, we know that the slope of line joining the points (a, b) and (c, d) is given by

$\rm\implies \:\boxed{\tt{ Slope \: of \: line \: = \: \frac{d - b}{c - a} \: }}$

So, using this the slope of line joining A(3, 4) and B(-3, -2) is given by

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

We know,

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

Thus,

$\bf\implies \:Slope \: of \: line \: l \: = \: - \: 1$

Now, We know that,

Equation of line which passes through the point (a, b) and having slope m is given by y - b = m(x - a).

So, Equation of perpendicular bisector l which passes through the point (0, 1) and having slope - 1, is given by

$\rm :\longmapsto\:y - 1 = - 1(x - 0)$

$\rm :\longmapsto\:y - 1 = - x$

$\bf :\longmapsto\:x + y - 1 = 0$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

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.