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#7.3

Find the ratio in which the line 2x + 3y – 5 = 0 divides the line segment joining the points (8, –9) and (2, 1). Also find the coordinates of the point of division.

Solution

$Section\, formula\left ( \frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}},\frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$

Let point p(x, y) divides the line segment joining the points A(8, –9) and B(2, 1) in ratio k : 1.
(x₁, y₁) = (8, -9) (x₂, y₂) = (2, 1)
m₁ : m2 = k:1
using section formula we have
$P\left ( x,y \right )= \left [ \frac{2k+8}{k+1},\frac{k-9}{k+1} \right ]$
Given equation is 2x + 3y – 5 = 0 …(2)
Put values of x and y in eqn. (2)
$2\left [ \frac{2k+8}{k+1} \right ]+3\left [ \frac{k-9}{k+1} \right ]-5= 0$

2(2k + 3) + 3(k – 9) – 5(k + 1) = 0
4k + 16 + 3k – 27 – 5k – 5 = 0
2k – 16 = 0
k = 8
Hence, p divides the line in ration 8 : 1.
Put k = 5 in eqn. (1)
$\left ( x,y \right )= \left [ \frac{2\left ( 8 \right )}{8+1},\frac{8-9}{8+1} \right ]$
$= \left ( \frac{16+8}{9},\frac{-1}{9} \right )= \left ( \frac{24}{9} ,\frac{-1}{9}\right )= \left ( \frac{8}{3},\frac{-1}{9} \right )$
Required point is $P\left ( \frac{8}{3} ,\frac{-1}{9}\right )$

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#7.3

Find the values of k if the points A(k + 1, 2k), B(3k, 2k + 3) and C (5k –1, 5k)are collinear.

Solution. If points A(k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are collinear then area of triangle is equal to zero
$\frac{1}{2}\left [ x_{1}\left ( y_{2}-y_{3} \right )+x_{2}\left ( y_{3}-y_{1} \right ) +x_{3}\left ( y_{1}-y_{2} \right )\right ]= 0$
$\left [ x_{1}\left ( y_{2}-y_{3} \right )+x_{2}\left ( y_{3}-y_{1} \right ) +x_{3}\left ( y_{1}-y_{2} \right )\right ]= 0$
[(k + 1) (2k + 3 – 5k) + 3k(5k – 2k) + (5k – 1)(2k – 2k – 3)] = 0
[(k + 1) (3 – 3k) + 3k(3k) + 5(k – 1) (–3)] = 0
[3k – 3k² + 3 – 3k + 9k² – 15k + 3] = 0
6k² – 15k + 6 = 0
6k² – 12k – 3k + 6 = 0
6k(k – 2) – 3(k – 2) = 0
(k – 2)(6k – 3 )
$k= 2,k= \frac{3}{6}$
$= \frac{1}{2}$
Hence, values of k are 2, 1/2

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#7.3

Find the coordinates of the point R on the line segment joining the points P(–1, 3) and Q(2, 5) such that $PR= \frac{3}{5}PQ$

Solution

$section\, formula \left ( \frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}},\frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$
According to question let R = (x, y) and PR = $\frac{3}{5}$ PQ
$\frac{PQ}{PR}= \frac{3}{5}$
R lies on PQ $\therefore$ PQ = PR +RQ

$\frac{PR+PQ}{PR}= \frac{5}{3}$
On dividing separately we get
$1+\frac{RQ}{PR}= \frac{5}{3}$
$\frac{RQ}{PR}= \frac{5}{3}-1= \frac{2}{3}$
$\Rightarrow PR:RQ= 3:2$
Hence, R divides PQ in ratio 3 : 2 using section formula we have
(x₁, y₁) = (-1, 3) (x₂, y₂) = (2, 5)
m₁ = 3, m₂ = 2
$R\left ( x,y \right )= \left ( \frac{3\left ( 2 \right )+2\left ( -1 \right )}{3+2} ,\frac{3\left ( 5 \right )+2\left ( 3 \right )}{3+2}\right )$
$R\left ( x,y \right )= \left ( \frac{6-2}{5},\frac{15+6}{5} \right )$
$R\left ( x,y \right )= \left ( \frac{4}{5},\frac{21}{5} \right )$
Here co- ordinates of R is $\left ( \frac{4}{5},\frac{21}{5} \right )$

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#7.3

The points A (2, 9), B (a, 5) and C (5, 5) are the vertices of a triangle ABC right angled at B. Find the values of a and hence the area of $\triangle$ ABC.

Solution.

$distance\, formula= \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}$
$\triangle$ ABC is a right angle triangle by using Pythagoras theorem we have
(AC)² = (BC)² + (AB)²
[(5 – 2)² + (5 - 9)²] = [(2 – a)²+ (9 – 5)²] + [(a – 5)² + (5 + 5)²]
(3)² + (–4)² = 4 + a² – 4a + (4)² + a² + 25 – 10a
9 + 16 = 4 + a² – 4a + 16 + a² + 25 – 10a
25 = 2a² – 14a + 45
2a² – 14a + 45 – 25 = 0
2a² – 14a + 20 = 0
Dividing by 2 we have
a² – 7a + 10 = 0
a² – 5a - 2a + 10 = 0
a(a – 5) – 2(a – 5) = 0
(a – 5) (a – 2) = 0
a = 5, a = 2
a = 5 is not possible because if a = 5 then point B and C coincide.
$\therefore$ a = 2
$Area \, of \, triangle = \frac{1}{2}\left [ x_{1} \left ( y_{2} -y_{3}\right )+x_{2} \left ( y_{3} -y_{1}\right )+x_{3} \left ( y_{1} -y_{2}\right )\right ]$
$= \frac{1}{2}\left [ 2\left ( 5-5 \right )+2\left ( 5-9 \right )+5\left ( 9-5 \right )\right ]$
$= \frac{1}{2}\left [ 0-8+20\right ]$
$= \frac{1}{2}\left [ 16\right ]$
= 8 sq. units.

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#7.3

$\text{If }$ $D\left ( \frac{-1}{2},\frac{5}{2} \right )$ $, E (7, 3) \ and$ $F\left ( \frac{7}{2},\frac{7}{2} \right )$ are the midpoints of sides of $\Delta ABC$ find the area of the $\Delta ABC$

Solution.

D is mid-point of AB using mid-point formula we get:
$-\frac{1}{2}= \frac{x_{1}+x_{2}}{2}$ $\frac{5}{2}= \frac{y_{1}+y_{2}}{2}$

x₁ + x₂ = –1 ……(1) 5 = y₁ + y₂ .…(2)

E is mid-point of BC using mid-point formula we get:
$\frac{x_{2}+x_{3}}{2}= 7$ $\frac{y_{2}+y_{3}}{2}= 3$

x₂ + x₃ = 14 …..(3) y₂ + y₃ = 6 ….(4)
F is mid-point of AB using mid-point formula we get:
$\frac{x_{1}+x_{3}}{2}= \frac{7}{2}$ $\frac{y_{1}+y_{3}}{2}= \frac{7}{2}$

x₁ + x₃ = 7       ….(5)               y₁ + y₃ = 7       …..(6)
Simplifying the above equations for values of x₁, y₁, x₂, y₂, x₃ and y₃
x₁ + x₂ = –1
x₃ + x₃ = 14               {using (1) and (3)}
– –   –
x₁ – x3 = –15   ….(7)
Use (5) and (7) we get
x₁ + x₃ = 7
x₁– x₃ = –15
2x₁ = –8
x₁ = –4
put x₁ = 4 in (1) we get
–4 + x₂ = –1
X₂ = –1 + 4 = 3
Punt x₂ = 3 in (3) we get
3 + x₃ = 14
x₃ = 11
Using equation (2) and (4) we get
y₁ + y₂ = 5
y₃ + y₃ = 6
- -   –
y₁ - y₃ = –1      ….(8)
Adding equation (6) and (8) we get
y₁ + y₃ = 7
y₁ – y₃ = –1
-------------------
2y₁ = 6
y₁ = 3
Put y₁ = 3 in eqn. (2)
5 – 3 = y₂
2 = y₂
Put y₂ = 2 in eqn. (4)
2 + y₃ = 6
y₃ = 6 – 2
y₃ = 4
Hence, x₁ = –4             y1 = 3
x₂ = 3               y₂ = 2
x₃ = 11             y₃ = 4
A = (x1, y₁) = (–4, 3), B = (x₂, y₂) = (3, 2), C = (x₃, y₃) = (11, 4)
$Area \, of \, triangle = \frac{1}{2}\left [ x_{1} \left ( y_{2} -y_{3}\right )+x_{2} \left ( y_{3} -y_{1}\right )+x_{3} \left ( y_{1} -y_{2}\right )\right ]$
$= \frac{1}{2}\left [ \left ( -4 \right )\left ( 2-4 \right )+\left ( 3 \right )\left ( 4-3 \right ) +11\left ( 3-2 \right )\right ]$
$= \frac{1}{2}\left [ \left ( -4 \right )\left ( -2 \right )+3\left ( 1 \right )+11\left ( 1 \right ) \right ]$
$= \frac{1}{2}\left [ 8+3+11 \right ]$
$= \frac{1}{2}\left [ 22 \right ]$
= 11 sq. units

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#7.3

The line segment joining the points A (3, 2) and B (5,1) is divided at the point P in the ratio 1:2 and it lies on the line 3x – 18y + k = 0. Find the value of k.

Solution:

Here points A(3, 2) and B(5, 1) is divided at the point P in the ratio 1 : 2
(x₁, y₁) = (3, 2)            (x₂, y₂) = (5, 1)
m₁ = 1, m₂ = 2
By section formula we have
$P\left ( x,y \right )= \left [ \frac{1\times 5+2\times 3}{1+2},\frac{1\times 1+2\times 2}{1+2} \right ]$
                 $= \left [ \frac{5+6}{3},\frac{1+4}{3} \right ]$
                 $= \left [ \frac{11}{3},\frac{5}{3} \right ]$
Line is 3x – 18y + k = 0 ......(1)
$Put\, point\, P\left ( \frac{11}{3},\frac{5}{3} \right ) in \left ( 1 \right )$
$Put\, x= \frac{11}{3},y= \frac{5}{3}$
$3\left ( \frac{11}{3} \right )-18\left ( \frac{5}{3} \right )+k= 0$

11 – 30 + k = 0
–19 + k = 0
k = 19

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#7.3

The centre of a circle is (2a, a – 7). Find the values of a if the circle passes through the point (11, –9) and has diameter $10\sqrt{2}$ units.

Solution

$distance\: formula= \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}$
Given points area A(2a, a –7) and B(11, –9)
$Diameter= 10\sqrt{2}$
$Radius= \frac{10\sqrt{2}}{2}= 5\sqrt{2}$
$Distance= 5\sqrt{2}$
(x₁, y₁) = (2a, a - 7) (x₂, y₂) = (11, -9)
$= \sqrt{\left ( 11-2a \right )^{2}+\left ( -9-a+7 \right )^{2}}= \left ( 5\sqrt{2} \right )$

Squaring both sides
$\left ( 11-2a \right )^{2}+\left ( -2-a \right )^{2}= \left ( 5\sqrt{2} \right )^{2}$
121 + 4a² – 44a + 4 + a² + 4a = 50
502 – 40a + 125 – 50 = 0
5a² – 40a + 75 = 0
Dividing by 5 we get
a² – 8a + 15 = 0
a² – 5a - 3a + 15 = 0
a(a – 5) – 3(a – 5) = 0
(a – 5) (a – 3) = 0
a = 5, 3

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#7.3

If (a, b) is the mid-point of the line segment joining the points A (10, –6) and B (k, 4) and a – 2b = 18, find the value of k and the distance AB.

Solution

$mid-point\, formula\, x= \frac{x_{1}+x_{2}}{2},y= \frac{y_{1}+y_{2}}{2}$
Point P (a, b) divide A(10, –6) and B(k, 4) in two equal parts.
$a= \frac{10+k}{2}$ $b= \frac{-6+4}{2}= \frac{-2}{2}= -1$
Given : a – 2b = 18
Put b = –1
a – 2(–1) = 18
a = 18 – 12
a = 16
$Now, a= \frac{10+k}{2}$
$16= \frac{10+k}{2}$
32 = 10 + k
32 – 10 = k
22 = k
$\therefore$ A (10, –6), B(22, 4)
$AB= \sqrt{\left ( 22-10 \right )^{2}+\left ( 4+6 \right )^{2}}$
$= \sqrt{\left ( 12 \right )^{2}+\left ( 10 \right )^{2}}$
$= \sqrt{144+100}$
$= \sqrt{244}$
$=2 \sqrt{61}$

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#7.3

If P(9a – 2, –b) divides line segment joining A(3a + 1, –3) and B(8a, 5)in the ratio 3: 1, find the values of a and b.

Solution

Point P(9a – 2, –b) divides line segment joining the points A(3a + 1, –3) and B(8a, 5) in ration 3 : 1.
(x₁, y₁) = (3a+1, -3)                 (x₂, y₂) = (8a, 5)
m₁ = 3, m₂ = 1
Using section formula we have
$\left ( 9a-2,-b \right )= \left [ \frac{3\left ( 8a \right )+1\left ( 3a+1 \right )}{3+1} ,\frac{3\left ( 5 \right )+1\left ( -3 \right )}{3+1}\right ]$
$\left ( 9a-2,-b \right )= \left [ \frac{24a+3a+1}{4},\frac{15-3}{4}\right ]$
$\left ( 9a-2,-b \right )= \left [ \frac{27a+1}{4},\frac{12}{4}\right ]$
Equate left-hand side and the right-hand side we get
$9a-2= \frac{27a+1}{4}$               $-b= \frac{12}{4}$

36a – 8 = 27a + 1 –b = 3
9a = 9 b = –3
$a= \frac{9}{9}$
a = 1

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#7.3

Find the ratio in which the point $P\left ( \frac{3}{4},\frac{5}{12} \right )$ divides the line segment joining the points $A\left ( \frac{1}{2},\frac{3}{2} \right )$ and $B (2, -5)$ .

Solution

$P\left ( X,Y \right )= \left ( \frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}},\frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$

Let the point $P\left ( \frac{3}{4},\frac{5}{12} \right )$ divides the line segment joining the points $A\left ( \frac{1}{2},\frac{3}{2} \right )$ and B(2, –5) in the ration k : 1.
(x₁, y₁) = (1/2, 3/2) (x₂, y₂) = (2, -5)
m₁ = k, m₂ = 1
Using section formula we have
$\left ( \frac{3}{4},\frac{5}{12} \right )= \left [ \frac{2k+1/2}{k+1} ,\frac{-5k+3/2}{k+1}\right ]$
$\left ( \frac{3}{4},\frac{5}{12} \right )= \left [ \frac{4k+1}{2k+1},\frac{-10k+3}{2k+2} \right ]$
$\frac{4k+1}{2k+1}= \frac{3}{4}$ $\frac{-10k+3}{2k+2}= \frac{5}{12}$

16k + 4 = 6k + 6 –120k + 36 = 10k + 10
16k – 6k = 6 – 4 –130k = –26
$k= \frac{2}{10}= \frac{1}{5}$
Therefore the required ratio is 1 : 5
i.e. 1 : 5

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Find the ratio in which the line 2x + 3y – 5 = 0 divides the line segment joining the points (8, –9) and (2, 1). Also find the coordinates of the point of division.

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The points A (2, 9), B (a, 5) and C (5, 5) are the vertices of a triangle ABC right angled at B. Find the values of a and hence the area of ABC.

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JEE Main high-scoring chapters and topics

are the midpoints of sides of find the area of the

Find the coordinates of the point R on the line segment joining the points P(–1, 3) and Q(2, 5) such that $PR= \frac{3}{5}PQ$

The points A (2, 9), B (a, 5) and C (5, 5) are the vertices of a triangle ABC right angled at B. Find the values of a and hence the area of $\triangle$ ABC.

$\text{If }$ $D\left ( \frac{-1}{2},\frac{5}{2} \right )$ $, E (7, 3) \ and$ $F\left ( \frac{7}{2},\frac{7}{2} \right )$ are the midpoints of sides of $\Delta ABC$ find the area of the $\Delta ABC$

The centre of a circle is (2a, a – 7). Find the values of a if the circle passes through the point (11, –9) and has diameter $10\sqrt{2}$ units.

Find the ratio in which the point $P\left ( \frac{3}{4},\frac{5}{12} \right )$ divides the line segment joining the points $A\left ( \frac{1}{2},\frac{3}{2} \right )$ and $B (2, -5)$ .