Answer

Verified

451.8k+ views

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.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Difference Between Plant Cell and Animal Cell

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Change the following sentences into negative and interrogative class 10 english CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Select the word that is correctly spelled a Twelveth class 10 english CBSE