Find the projection of the point (1,0) on the line joining the points P(−1,2) and Q(5,4).
Answer
362.4k+ 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.
Last updated date: 29th Sep 2023
•
Total views: 362.4k
•
Views today: 3.62k
Recently Updated Pages
What is the Full Form of DNA and RNA

What are the Difference Between Acute and Chronic Disease

Difference Between Communicable and Non-Communicable

What is Nutrition Explain Diff Type of Nutrition ?

What is the Function of Digestive Enzymes

What is the Full Form of 1.DPT 2.DDT 3.BCG

Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Summary of the poem Where the Mind is Without Fear class 8 english CBSE

Difference Between Plant Cell and Animal Cell

What is the basic unit of classification class 11 biology CBSE

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

One cusec is equal to how many liters class 8 maths CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Give 10 examples for herbs , shrubs , climbers , creepers
