# Find the equation of the perpendicular bisector of a straight line segment joining the points$(3,4)$,$(-1,2)$.

Last updated date: 25th Mar 2023

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Hint: Name the points and perpendicular bisector. Find the midpoint and slope of $(3,4)$ and $(-1,2)$. Put ${{m}_{1}}{{m}_{2}}=-1$ and you will get the slope of the perpendicular bisector. After that use slope point form and you will get the equation.

A perpendicular bisector can be defined as a line segment that intersects another line perpendicularly and divides it into two equal parts.

Two lines are said to be perpendicular to each other when they intersect in such a way that they form $90{}^\circ $ with each other. And, a bisector divides a line into two equal halves.

Thus, a perpendicular bisector of a line segment $AB$ implies that it intersects $AB$ at $90{}^\circ $ and cuts it into two halves.

Properties of a Perpendicular Bisector

1) It divides $AB$ into two equal halves or bisects it.

2) It makes right angles with (or is perpendicular to) $AB$.

3) Every point in the perpendicular bisector is equidistant from point $A$ and $B$.

While working with practical geometry, you will often find the application of perpendicular bisectors; say when you are asked to draw an isosceles triangle, or when you have to determine the center of a circle, etc.

So in question, there are two points, so let us consider the points as $A(3,4)$ and $B(-1,2)$.

So $AB$ is the segment and we want to find the equation of perpendicular bisector i.e. to find the equation of line $P$ perpendicular to the segment $AB$.

So as it is a perpendicular bisector, so the point $C$ is the midpoint of the segment $AB$.

So point $C$ should be $(a,b)$, from which line $P$ passes.

So the point $C$ is the midpoint of the segment $AB$.

$(a,b)=\left( \dfrac{3+(-1)}{2},\dfrac{4+2}{2} \right)=\left( 1,3 \right)$

So we get a point $C(1,3)$.

Now we have to find the slope of the line $P$.

For that, we should find the slope of the segment $AB$.

So let${{m}_{1}}$ be the slope of the segment $AB$.

${{m}_{1}}=\dfrac{2-4}{-1-3}=\dfrac{-2}{-4}=\dfrac{1}{2}$

So ${{m}_{1}}=\dfrac{1}{2}$,

Now we know if a line is perpendicular to segment, so the product of their slopes is equal to $-1$.

So let slope of the line $P$ be ${{m}_{2}}$.

So we get,

${{m}_{1}}{{m}_{2}}=-1$

$\dfrac{1}{2}{{m}_{2}}=-1$

So we get ${{m}_{2}}$.

${{m}_{2}}=-2$,

So we have got the slope of a line $P$and also know that it is the perpendicular bisector of the line segment $AB$.

So it passes through the point $C$.

So now we have a slope and a point so we can use the slope point form to find the equation of the perpendicular bisector.

So general form of slope point form is, if the point is $({{x}_{1}},{{y}_{1}})$ from which the line passes, and its slope $m=-2$, then, the line is given by :

$y-{{y}_{1}}=m(x-{{x}_{1}})$

So substituting for $m,{{x}_{1}},{{y}_{1}}$, we get,

$y-3=\left( -2 \right)(x-1)$

So simplifying in a simple manner we get,

$\begin{align}

& y-3=-2x+2 \\

& 2x+y=5 \\

\end{align}$

So we get the equation of perpendicular bisector i.e. line $P$ of the straight line segment joining the points $(3,4)$,$(-1,2)$ is $2x+y=5$.

Note: Read the question carefully. Be familiar with the properties. Don’t confuse yourself with ${{m}_{1}}{{m}_{2}}=-1$ this one. Also, you should know the general form of slope point form $y-{{y}_{1}}=m(x-{{x}_{1}})$. While substituting do not confuse with the points. Also while finding the slope be careful of the substitutions.

A perpendicular bisector can be defined as a line segment that intersects another line perpendicularly and divides it into two equal parts.

Two lines are said to be perpendicular to each other when they intersect in such a way that they form $90{}^\circ $ with each other. And, a bisector divides a line into two equal halves.

Thus, a perpendicular bisector of a line segment $AB$ implies that it intersects $AB$ at $90{}^\circ $ and cuts it into two halves.

Properties of a Perpendicular Bisector

1) It divides $AB$ into two equal halves or bisects it.

2) It makes right angles with (or is perpendicular to) $AB$.

3) Every point in the perpendicular bisector is equidistant from point $A$ and $B$.

While working with practical geometry, you will often find the application of perpendicular bisectors; say when you are asked to draw an isosceles triangle, or when you have to determine the center of a circle, etc.

So in question, there are two points, so let us consider the points as $A(3,4)$ and $B(-1,2)$.

So $AB$ is the segment and we want to find the equation of perpendicular bisector i.e. to find the equation of line $P$ perpendicular to the segment $AB$.

So as it is a perpendicular bisector, so the point $C$ is the midpoint of the segment $AB$.

So point $C$ should be $(a,b)$, from which line $P$ passes.

So the point $C$ is the midpoint of the segment $AB$.

$(a,b)=\left( \dfrac{3+(-1)}{2},\dfrac{4+2}{2} \right)=\left( 1,3 \right)$

So we get a point $C(1,3)$.

Now we have to find the slope of the line $P$.

For that, we should find the slope of the segment $AB$.

So let${{m}_{1}}$ be the slope of the segment $AB$.

${{m}_{1}}=\dfrac{2-4}{-1-3}=\dfrac{-2}{-4}=\dfrac{1}{2}$

So ${{m}_{1}}=\dfrac{1}{2}$,

Now we know if a line is perpendicular to segment, so the product of their slopes is equal to $-1$.

So let slope of the line $P$ be ${{m}_{2}}$.

So we get,

${{m}_{1}}{{m}_{2}}=-1$

$\dfrac{1}{2}{{m}_{2}}=-1$

So we get ${{m}_{2}}$.

${{m}_{2}}=-2$,

So we have got the slope of a line $P$and also know that it is the perpendicular bisector of the line segment $AB$.

So it passes through the point $C$.

So now we have a slope and a point so we can use the slope point form to find the equation of the perpendicular bisector.

So general form of slope point form is, if the point is $({{x}_{1}},{{y}_{1}})$ from which the line passes, and its slope $m=-2$, then, the line is given by :

$y-{{y}_{1}}=m(x-{{x}_{1}})$

So substituting for $m,{{x}_{1}},{{y}_{1}}$, we get,

$y-3=\left( -2 \right)(x-1)$

So simplifying in a simple manner we get,

$\begin{align}

& y-3=-2x+2 \\

& 2x+y=5 \\

\end{align}$

So we get the equation of perpendicular bisector i.e. line $P$ of the straight line segment joining the points $(3,4)$,$(-1,2)$ is $2x+y=5$.

Note: Read the question carefully. Be familiar with the properties. Don’t confuse yourself with ${{m}_{1}}{{m}_{2}}=-1$ this one. Also, you should know the general form of slope point form $y-{{y}_{1}}=m(x-{{x}_{1}})$. While substituting do not confuse with the points. Also while finding the slope be careful of the substitutions.

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