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

Answer

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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.

Last updated date: 29th Sep 2023

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