# Find the projection of the point (1,0) on the line joining the points P(−1,2) and Q(5,4).

Last updated date: 26th Mar 2023

•

Total views: 307.5k

•

Views today: 8.84k

Answer

Verified

307.5k+ views

Hint: Let A(1,0) be the given point and M(h,k) be the projection on the line PQ. Equate the slopes of

PQ and MQ to get one equation. AM and PQ are perpendicular. So use the relation between slopes

of perpendicular lines to get the second equation. Solve these equations to get M(h,k).

In this question, we need to find the projection of the point (1,0) on the line joining the

points P(−1,2) and Q(5,4).

Let A(1,0) be the given point whose projection is to be evaluated.

Also, let M(h,k) be the foot of the perpendicular drawn from A(1,0) to the line joining the

points P(−1,2) and Q(5,4).

This means that M(h,k) lies on the line joining the points P(−1,2) and Q(5,4).

So, the slope of PQ will be equal to the slope of MQ.

We know that the slope of a line joining the points (a,b) and (c,d) is:

$m=\dfrac{\left( d-b \right)}{\left( c-a \right)}$

Now, we will find the slope of the line joining the points P(−1,2) and Q(5,4).

${{m}_{PQ}}=\dfrac{\left( 4-2 \right)}{\left( 5+1 \right)}$$=\dfrac{2}{6}$

${{m}_{PQ}}=\dfrac{1}{3}$

…(1)

Similarly, we will find the slope of the segment joining the points M(h,k) and Q(5,4).

${{m}_{MQ}}=\dfrac{\left( 4-k \right)}{\left( 5-h \right)}$

…(2)

Now, since the slope of PQ will be equal to the slope of MQ, we will equate (1) and (2).

$\dfrac{1}{3}=\dfrac{\left( 4-k \right)}{\left( 5-h \right)}$

$12-3k=5-h$

$h-3k+7=0$ …(3)

Now, we know that AM is perpendicular to the line PQ.

We also know that if two lines AB and CD with slopes ${{m}_{AB}}$ and ${{m}_{CD}}$

respectively, are perpendicular to each other, then:

${{m}_{AB}}\cdot {{m}_{CD}}=-1$

We will now find the slope of the line joining A(1,0) and M(h,k)

${{m}_{AM}}=\dfrac{\left( k-0 \right)}{\left( h-1 \right)}$

${{m}_{AM}}=\dfrac{k}{\left( h-1 \right)}$

…(4)

We also know from (1) that ${{m}_{PQ}}=\dfrac{1}{3}$ .

Now, since AM is the perpendicular to the line PQ, we have the following:

${{m}_{AM}}\cdot {{m}_{PQ}}=-1$

$\dfrac{1}{3}\times \dfrac{k}{\left( h-1 \right)}=-1$

$k=3-3h$

$3h+k-3=0$ …(5)

We have two linear equations in two variables: (3) and (5).

We will solve these to find the values of h and k.

From (3), we have $h=3k-7$ . We will substitute this in (5) to get:

$3\left( 3k-7 \right)+k-3=0$

$9k-21+k-3=0$

$10k=24$

$k=\dfrac{12}{5}$

We will now put this value in $h=3k-7$ to get the value of h.

$h=3\times \dfrac{12}{5}-7$

$h=\dfrac{36}{5}-\dfrac{35}{5}$

$h=\dfrac{1}{5}$

Hence, we have $h=\dfrac{1}{5}$ and $k=\dfrac{12}{5}$ .

So, the projection of the point (1,0) on the line joining the points P(−1,2) and Q(5,4) is M $\left(

\dfrac{1}{5},\dfrac{12}{5} \right)$ .

Note: Another method to solve this question can be the following. Find the equation of the line PQ

using the formula $y-{{y}_{1}}={{m}_{PQ}}\left( x-{{x}_{1}} \right)$ . Then using the formula

$\dfrac{\left( x-{{x}_{1}} \right)}{a}=\dfrac{\left( y-{{y}_{1}} \right)}{b}=\dfrac{-\left(

a{{x}_{1}}+b{{y}_{1}}+c \right)}{{{a}^{2}}+{{b}^{2}}}$ find the projection of A on PQ.

PQ and MQ to get one equation. AM and PQ are perpendicular. So use the relation between slopes

of perpendicular lines to get the second equation. Solve these equations to get M(h,k).

In this question, we need to find the projection of the point (1,0) on the line joining the

points P(−1,2) and Q(5,4).

Let A(1,0) be the given point whose projection is to be evaluated.

Also, let M(h,k) be the foot of the perpendicular drawn from A(1,0) to the line joining the

points P(−1,2) and Q(5,4).

This means that M(h,k) lies on the line joining the points P(−1,2) and Q(5,4).

So, the slope of PQ will be equal to the slope of MQ.

We know that the slope of a line joining the points (a,b) and (c,d) is:

$m=\dfrac{\left( d-b \right)}{\left( c-a \right)}$

Now, we will find the slope of the line joining the points P(−1,2) and Q(5,4).

${{m}_{PQ}}=\dfrac{\left( 4-2 \right)}{\left( 5+1 \right)}$$=\dfrac{2}{6}$

${{m}_{PQ}}=\dfrac{1}{3}$

…(1)

Similarly, we will find the slope of the segment joining the points M(h,k) and Q(5,4).

${{m}_{MQ}}=\dfrac{\left( 4-k \right)}{\left( 5-h \right)}$

…(2)

Now, since the slope of PQ will be equal to the slope of MQ, we will equate (1) and (2).

$\dfrac{1}{3}=\dfrac{\left( 4-k \right)}{\left( 5-h \right)}$

$12-3k=5-h$

$h-3k+7=0$ …(3)

Now, we know that AM is perpendicular to the line PQ.

We also know that if two lines AB and CD with slopes ${{m}_{AB}}$ and ${{m}_{CD}}$

respectively, are perpendicular to each other, then:

${{m}_{AB}}\cdot {{m}_{CD}}=-1$

We will now find the slope of the line joining A(1,0) and M(h,k)

${{m}_{AM}}=\dfrac{\left( k-0 \right)}{\left( h-1 \right)}$

${{m}_{AM}}=\dfrac{k}{\left( h-1 \right)}$

…(4)

We also know from (1) that ${{m}_{PQ}}=\dfrac{1}{3}$ .

Now, since AM is the perpendicular to the line PQ, we have the following:

${{m}_{AM}}\cdot {{m}_{PQ}}=-1$

$\dfrac{1}{3}\times \dfrac{k}{\left( h-1 \right)}=-1$

$k=3-3h$

$3h+k-3=0$ …(5)

We have two linear equations in two variables: (3) and (5).

We will solve these to find the values of h and k.

From (3), we have $h=3k-7$ . We will substitute this in (5) to get:

$3\left( 3k-7 \right)+k-3=0$

$9k-21+k-3=0$

$10k=24$

$k=\dfrac{12}{5}$

We will now put this value in $h=3k-7$ to get the value of h.

$h=3\times \dfrac{12}{5}-7$

$h=\dfrac{36}{5}-\dfrac{35}{5}$

$h=\dfrac{1}{5}$

Hence, we have $h=\dfrac{1}{5}$ and $k=\dfrac{12}{5}$ .

So, the projection of the point (1,0) on the line joining the points P(−1,2) and Q(5,4) is M $\left(

\dfrac{1}{5},\dfrac{12}{5} \right)$ .

Note: Another method to solve this question can be the following. Find the equation of the line PQ

using the formula $y-{{y}_{1}}={{m}_{PQ}}\left( x-{{x}_{1}} \right)$ . Then using the formula

$\dfrac{\left( x-{{x}_{1}} \right)}{a}=\dfrac{\left( y-{{y}_{1}} \right)}{b}=\dfrac{-\left(

a{{x}_{1}}+b{{y}_{1}}+c \right)}{{{a}^{2}}+{{b}^{2}}}$ find the projection of A on PQ.

Recently Updated Pages

If ab and c are unit vectors then left ab2 right+bc2+ca2 class 12 maths JEE_Main

A rod AB of length 4 units moves horizontally when class 11 maths JEE_Main

Evaluate the value of intlimits0pi cos 3xdx A 0 B 1 class 12 maths JEE_Main

Which of the following is correct 1 nleft S cup T right class 10 maths JEE_Main

What is the area of the triangle with vertices Aleft class 11 maths JEE_Main

KCN reacts readily to give a cyanide with A Ethyl alcohol class 12 chemistry JEE_Main

Trending doubts

What was the capital of Kanishka A Mathura B Purushapura class 7 social studies CBSE

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Tropic of Cancer passes through how many states? Name them.

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

Name the Largest and the Smallest Cell in the Human Body ?