Q. 1: Determine the ratio, in which the line \(2x + y – 4 = 0 \)
divides the line segment joining the points \(A(2, -2)\) and \(B(3, 7)\).
Solution
Let the line \(2x + y – 4 = 0 \) divide the line \(\;AB\)
at \(\;Q\) in the ratio of \(k:1\), i.e. \(m:n = k:1\).
\begin{align}
& m=k,\; n=1, \\
& x_1= 2,\; y_1=-2,\; x_2=3,\; y_2=7 \\
\end{align}
So, coordinates of \(Q(x, y)\) are
\[ \left(\,
\frac { k \cdot 3 + 1 \cdot 2 } { k + 1},
\frac { k \cdot 7 + 1 \cdot (-2) } { k + 1} \,
\right)
\]
\[\left(\frac{3k + 2}{k + 1},\, \frac{7k - 2}{k + 1} \, \right)\]
This point
\(\left(\frac{3k + 2}{k + 1},\, \frac{7k - 2}{k + 1} \, \right)\)
lies on both the lines \(AB\) and line \(2x + y – 4 = 0\).
So, putting
\(\frac{3k + 2} {k + 1}\)
for \(x\) and
\(\frac{7k - 2} {k + 1}\)
for \(y\) in the given equation,
\begin{align}
& 2 \times \left( \frac{3k+2}{k+1} \right) + \left( \frac{7k-2}{k+1} \right) -4 = 0 \\\\
& \Rightarrow \frac{6k+4}{k+1} + \frac{7k-2}{k+1} - 4 = 0 \\\\
& \Rightarrow \frac{6k+4+7k-2}{k+1} = 4 \\\\
& \Rightarrow \frac{13k+2}{k+1} = 4 \\\\
& \Rightarrow 13k+2 = 4(k+1) \quad\color{gray}\text{[by cross multiplication]} \\\\
& \Rightarrow 13k+2 = 4k+4 \\\\
& \Rightarrow 13k-4k = 4-2 \\\\
& \Rightarrow 9k = 2 \\\\
& \therefore k = \frac29 \\\\
\end{align}
So, the required ratio is \(2:9\) Ans.
Q. 2: Find a relation between x and y,
if the points \((x, y),\; (1, 2)\) and \((7, 0)\) are collinear.
Solution
Since the given points are collinear, the area of the triangle formed by them is 0.
Area of a triangle w.r.t its co-ordinates of the vertices is:
\[\normalsize{\frac12 {\color{red} \{}
x_1(y_2 - y_3) \boldsymbol{\color{red} +} x_2(y_3 - y_1) \boldsymbol{\color{red} +} x_3(y_1 - y_2)
{\color{red} \}} }\]
Here we get
\begin{align}
& x_1=x,\; y_1=y, \\
& x_2= 1,\; y_2=2,\; x_3=7,\; y_3=0 \\
\end{align}
\[\frac12 {\color{red} \{}
x(2 - 0) \boldsymbol{\color{red} +} 1(0 - y) \boldsymbol{\color{red} +} 7(y - 2)
{\color{red} \}} \]
\[ \Rightarrow \frac12 \times \{x\cdot 2 + 1\cdot(-y)+7\cdot y - 7\cdot 2\} = 0\]
\[ \Rightarrow \frac12 \times \{2x - y + 7y - 14\} = 0\]
\[ \Rightarrow \frac12 \times \{2x + 6y - 14\} = 0\]
\[ \Rightarrow \frac12 \times 2\times(x + 3y - 7) = 0\]
\[ \Rightarrow x + 3y - 7 = 0\]
\[ \Rightarrow 3y = -x + 7 \]
\[ \Rightarrow y = \frac{-x + 7}3 \]
\[ \Rightarrow y = -\frac13 x + \frac73 \]
So, the required relation is \(\;y = -\frac13 x + \frac73\; \)
Ans.
Following is the straight line of the equation
Q. 3: Find the centre of a circle passing through
the points \((6, -6),\; (3, -7)\;\) and \(\;(3, 3)\).
Solution
Given three points \(A(6, -6),\; B(3, -7)\;\) and \(\;C(3, 3)\;\) are located
on the circumference of a circle.
Let \(\,O(x, y)\,\) be the center of this circle.
So, \(\;OA, OB\) and \(\;OC\) are the radii of the circle,
and \(\,OA=OB=OC\).
Formula of the distance between two points is:
For OA, as per
as per distance formula
,
\[x_1= x,\, y_1= y,\, x_2= 6,\, y_2= -6\]
\begin{align}
OA &= \sqrt{(6-x)^2 + (-6-y)^2} \\
&= \sqrt{6^2-2 \cdot 6 \cdot x + x^2 + (-6)^2 -2 \cdot(-6) \cdot y + y^2} \\
&= \sqrt{36 - 12x + x^2 + 36 + 12y + y^2} \\
&= \sqrt{x^2 + y^2 - 12x + 12y + 72 } \\
\end{align}
For OB,
\[x_1= x,\, y_1= y,\, x_2= 3,\, y_2= -7\]
\begin{align}
OB &= \sqrt{(3-x)^2 + (-7-y)^2} \\
&= \sqrt{3^2-2\cdot 3\cdot x + x^2 + (-7)^2 -2 \cdot(-7) \cdot y + y^2} \\
&= \sqrt{9 - 6x + x^2 + 49 + 14y + y^2} \\
&= \sqrt{x^2 + y^2 - 6x + 14y + 58 } \\\\
\end{align}
For OC,
\[x_1= x,\, y_1= y,\, x_2= 3,\, y_2= 3\]
\begin{align}
OC &= \sqrt{(3-x)^2 + (3-y)^2} \\
&= \sqrt{3^2-2\cdot 3\cdot x + x^2 + 3^2 -2 \cdot 3 \cdot y + y^2} \\
&= \sqrt{9 - 6x + x^2 + 9 - 6y + y^2} \\
&= \sqrt{x^2 + y^2 - 6x - 6y + 18 } \\\\
\end{align}
Because OA and OB are the radii of same circle, \(OA = OB\)
\begin{multline}
\therefore \sqrt{x^2 + y^2 - 12x + 12y + 72 } \\
= \sqrt{x^2 + y^2 - 6x + 14y + 58 } \\
\end{multline}
Taking square of both the sides,
\begin{multline}
\Rightarrow \cancel{x^2} + \cancel{y^2} - 12x + 12y + 72 \\
= \cancel{x^2} + \cancel{y^2} - 6x + 14y + 58 \\
\end{multline}
\begin{align}
&\Rightarrow - 12x + 6x + 12y - 14y = 58-72 \\
&\Rightarrow - 6x - 2y = -14 \\
&\Rightarrow -3x - y = -7 \quad \bbox{\text{... eqn(i)}}\\
\end{align}
Again, OA and OC are the radii of same circle, \(OA = OC\)
\begin{multline}
\therefore \sqrt{x^2 + y^2 - 12x + 12y + 72 } \\
= \sqrt{x^2 + y^2 - 6x - 6y + 18 } \\
\end{multline}
Taking square of both the sides,
\begin{multline}
\Rightarrow \cancel{x^2} + \cancel{y^2} - 12x + 12y + 72 \\
\shoveleft{\hspace{2cm} = \cancel{x^2} + \cancel{y^2} - 6x - 6y + 18} \\
\end{multline}
\begin{align}
&\Rightarrow - 12x + 6x + 12y + 6y = 18-72 \\
&\Rightarrow - 6x + 18y = -54 \\
&\Rightarrow - 3x + 9y = -27 \quad \bbox{\text{... eqn(ii)}}\\
\end{align}
Subtracting eqn(i) from eqn(ii),
\begin{align}
\cancel{-3x} + 9y &= -27 \\
\cancel{-3x} - y &= -7 \\
\end{align}
\[------------- \]
\[9y-(-y) = -27-(-7) \]
\[\Rightarrow 9y+y = -27+7 \]
\[\Rightarrow 10y = -20 \]
\[\therefore y = -2 \]
Putting \(\;y=-2\;\) in equation ... (i),
\begin{align}
&-3x - (-2) = -7 \\
&\Rightarrow -3x + 2 = -7 \\
&\Rightarrow -3x = -7-2 \\
&\Rightarrow -3x = -9 \\
&\therefore x = 3
\end{align}
So, the required coordinates of center O are \((3, -2)\)
Ans.
Following is the pictorial representation of the problem:
Q. 4: The two opposite vertices of a square are
\((-1, 2)\;\) and \((3, 2)\). Find the coordinates of the other two vertices.
Solution
Let ABCD be the square whose two opposite vertices are given as
\(A(-1, 2)\;\) and \(C(3, 2)\).
We have to find coordinates of the vertices \(B\) and \(D\).
Let the coordinates of \(B\) be \((x, y)\).
Step #1: We find \( AB, BC\;\) and \(AC \)
Step #2: with \(AB = BC\), we solve for x,
Step #3: with \(AB, BC\;\) and \(AC\), we solve for y.
Distance between \( \;A(-1, 2) \;\) and\(\; B(x, y)\)
Q. 5: The class X students school in krishnagar have been allotted
a rectangular plot of land for their gardening activity. Saplings of Gulmohar
are planted on the boundary at a distance of 1 m from each other. There is
triangular grassy lawn in the plot as shown in the figure. The students are
to sow seeds of flowering plants on the remaining area of the plot.
(i) Taking A as origin, find the coordinates of the vertices of the triangle.
(ii) What will be the coordinates of the vertices of ∆PQR, if C is the origin?
Also, calculate the areas of the triangles in these cases. What do you observe?
Solution
Let ABCD be the rectangular garden as above.
If we consider point A as the origin,
AD is the X-axis and AB is the Y-axis.
So coordinates of points
P, Q and R are \(P(4,6),\; Q(3,2)\;\) and \(\;R(6,5)\).
Here we get
\begin{align}
& x_1=4,\; y_1=6, \\
& x_2= 3,\; y_2=2,\; x_3=6,\; y_3=5 \\
\end{align}
We calculate the area of ΔPQR =
\[\frac12 \times {\color{red} \{}
4\times(2 - 5) \boldsymbol{\color{red} +} 3\times(5 - 6) \boldsymbol{\color{red} +} 6\times(6 - 2)
{\color{red} \}} \]
\[\frac12 \times \{
4 \times (-3) + 3 \times (-1) + 6 \times 4
\} \]
\[ = \frac12 \times \{-12-3+24\}\]
\[ = \frac12 \times 9\]
\[ = \frac92 \text{ sq. unit} \]
If we consider point \(C\) as the origin,
\(CB\) is the X-axis and \(CD\) is the Y-axis.
So coordinates of the points
\(P, Q, R\) are \(P(-12, -2),\; Q(-13, -6), \;R(-10, -3)\).
After shifting of the origin of the graph, we get
\begin{align}
& x_1=-12,\; y_1=-2, \\
& x_2= -13,\; y_2=-6,\; x_3=-10,\; y_3=-3 \\
\end{align}
Area of ΔPQR with those new coordinates
\[\Tiny{\frac12 {\color{red} \{}
-12(-6 - (-3)) \boldsymbol{\color{red} +} (-13)(-3 - (-2)) \boldsymbol{\color{red} +} (-10)(-2 - (-6))
{\color{red} \}} }\]
\[\small{\frac12 {\color{red} \{}
-12(-6 +3) -13(-3 +2) -10(-2 +6)
{\color{red} \}} }\]
\[\small{\frac12 {\color{red} \{}
-12(-3) -13(-1) -10 4) ;
{\color{red} \}} }\]
\[ = \frac12 \times (36+13-40)\]
\[ = \frac12 \times (49-40)\]
\[ = \frac12 \times 9\]
\[ = \frac92 \text{ sq. unit} \]
(i) So, when point A is the origin,
the required coordinates
are \(\;P(4, 6),\; Q(3, 2)\;\) and \(\;R(6, 5)\)
Ans.
(ii) When point C is the origin,
the required coordinates
are \(\;P(-12, -2),\; Q(-13, -6)\;\) and \(\;R(-10, -3)\)
Ans.
In both the cases, area of the triangle is \(\frac92\) sq unit.
Observation: The area does not change with change in the origin.
Ans.
Q. 6: The vertices of a ΔABC are
\(\;A(4, 6),\; B(1, 5)\;\) and \(\;C(7, 2)\).
A line is drawn to intersect sides AB and AC at D and E respectively,
such that \(\frac{AD}{AB}=\frac{AE}{AC}=\frac14\).
Calculate the area of the ΔADE and compare it with the area of ΔABC.
Solution
Step #1: Given \(\;AD:AB\), we find \(\;AD:DB \).
Step #2: Similarly, we find \(\;AE:EC\).
Step #3: With these ratios, find coordinate of \(\;D\;\) and \(\;E\;\).
Step #4: Calculate the area of triangles \(\;ADE\;\) and \(\;ABC\)
Step #5: Finally we compare the areas.
For ΔABC, vertices are
\begin{align}
& x_1=4,\; y_1=6, \\
& x_2= 1,\; y_2=5,\; x_3=7,\; y_3=2 \\
\end{align}
∴ Area of ΔABC
\[\frac12 \times {\color{red} \{}
4\times(5 - 2) \boldsymbol{\color{red} +} 1\times(2 - 6) \boldsymbol{\color{red} +} 7\times(6 - 5)
{\color{red} \}} \]
\[ = \frac12 \times (4 \cdot 3 + 1 \cdot(-4) + 7 \cdot 1) \]
\[ = \frac12 \times (12-4+7) \]
\[ = \frac{15}2 \text{ sq. unit} \]
∴ Ratios of areas of the triangles
\[\frac{\triangle ADE}{\triangle ABC} \]
\[ = \frac{15}{32} \div \frac{15}{2} \]
\[ = \frac{15}{32} \times \frac{2}{15} \]
\[ = \frac{1}{16} \]
\[ = 1:16 \]
∴ Area of ΔADE = \( \frac{15}{32} \text{ sq. unit} \)
and ratios of areas of ΔADE to ΔABC is \(1:16\)
Ans.
Q. 7: Let \(\;A(4, 2),\; B(6,5)\;\) and \(\;C(1, 4)\;\) be the vertices of ∆ABC. (i)
The median from \(\; A\) meets \(\; BC\) at \(\; D\). Find the coordinates of point \(\; D\). (ii)
Find the coordinates of the point \(\; P \) on \(\; AD \), such that \(\;AP: PD = 2: 1\). (iii)
Find the coordinates of points \(\; Q \) and \(\; R \) on medians \(\; BE, CF\;\) respectively
such that \(\;BQ:QE = 2:1\;\) and \(\;CR:RF = 2:1\). (iv) What do you observe? (v)
If \( A(x_1, y_1),\; B(x_2, y_2)\;\) and \(\;C(x_3, y_3) \) are the vertices of
\(\; ∆ABC \), find the coordinates of the centroid of the triangle.
Solution
Step #1: Given \(\;AD:AB\), we find \(\;AD:DB \).
Step #2: Similarly, we find \(\;AE:EC\).
Step #3: With these ratios, find coordinate of \(\;D\;\) and \(\;E\;\).
Step #4: Calculate the area of triangles \(\;ADE\;\) and \(\;ABC\)
Step #5: Finally we compare the areas.
Let us find coordinates of midpoints \(\; D \), \(\; E \) and \(\; F \) with the
midpoint formula:
\[\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \, \right)\]
Solution 7 (i):
Coordinates of point \(\; D \) (midpoint of \(\; BC \)) when
\[x_1= 6,\, y_1= 5,\, x_2= 1,\, y_2= 4\]
\[\left(\frac{6+1}{2}, \frac{5+4}{2} \, \right)\]
\[ = \left( \frac72, \frac92 \right) \]
Coordinates of point \(\; E \) (midpoint of \(\; AC \)) when
\[x_1= 4,\, y_1= 2,\, x_2= 1,\, y_2= 4\]
\[\left(\frac{4+1}{2}, \frac{2+4}{2} \, \right)\]
\[ = \left( \frac52, \frac62 \right) \]
Coordinates of point \(\; F \) (midpoint of \(\; AB \)) when
\[x_1= 4,\, y_1= 2,\, x_2= 6,\, y_2= 5\]
\[\left(\frac{4+6}{2}, \frac{2+5}{2} \, \right)\]
\[ = \left( \frac{10}2, \frac72 \right) \]
With coordinates of points \(\; A, D\;\) and given \(\;m:n=2:1\),
we find coordinates of point \(\; P \;\) by using the
section formula:
\[ \left(\,
\frac { m \cdot x_2 + n \cdot x_1 } { m + n},
\frac { m \cdot y_2 + n \cdot y_1 } { m + n} \,
\right)
\]
Solution 7 (ii):
Coordinates of point \(\; P \) when
\begin{align}
& m=2,\; n=1, \\
& x_1= 4,\; y_1=2,\; x_2=\frac72,\; y_2=\frac92 \\
\end{align}
\[ \left(\,
\frac { 2 \cdot \frac72 + 1 \cdot 4 } { 2 + 1},
\frac { 2 \cdot \frac92 + 1 \cdot 2 } { 2 + 1} \,
\right)
\]
\[\left(\frac{7 + 4}{3},\, \frac{9 + 2}{3} \, \right)\]
\[\left(\frac{11}{3},\, \frac{11}{3} \, \right)\]
Similarly, with coordinates of points \(\; B, E\;\) and given \(\;m:n=2:1\;\),
we find coordinates of point \(\; Q \).
Solution 7 (iii):
Coordinates of point \(\; Q \) when
\begin{align}
& m=2,\; n=1, \\
& x_1= 6,\; y_1=5,\; x_2=\frac52,\; y_2=\frac62 \\
\end{align}
\[ \left(\,
\frac { 2 \cdot \frac52 + 1 \cdot 6 } { 2 + 1},
\frac { 2 \cdot \frac62 + 1 \cdot 5 } { 2 + 1} \,
\right)
\]
\[\left(\frac{5 + 6}{3},\, \frac{6 + 5}{3} \, \right)\]
\[\left(\frac{11}{3},\, \frac{11}{3} \, \right)\]
Similarly, with coordinates of points \(\; C, F\;\) and given \(\;m:n=2:1\;\),
we find coordinates of point \(\; R \).
Coordinates of point \(\; R \) when
\begin{align}
& m=2,\; n=1, \\
& x_1= 1,\; y_1=4,\; x_2=\frac{10}2,\; y_2=\frac72 \\
\end{align}
\[ \left(\,
\frac { 2 \cdot \frac{10}2 + 1 \cdot 1 } { 2 + 1},
\frac { 2 \cdot \frac72 + 1 \cdot 4 } { 2 + 1} \,
\right)
\]
\[\left(\frac{10 + 1}{3},\, \frac{7 + 4}{3} \, \right)\]
\[\left(\frac{11}{3},\, \frac{11}{3} \, \right)\]
Solution 7 (iv):
Hence it can be observed that \(\; P, Q\) and \(\; R\) are the same point.
This point, which is
common to all the three medians,
divides each median in the ratio of \(\;2:1\).
This point \( \left(\frac{11}3, \; \frac{11}3 \right)\; \)
is called the centroid of the triangle.
Solution 7 (v):
Now, if the vertices of the triangle \(\; ABC\;\) are
\( A(x_1, y_1), \;B(x_2, y_2) \;\) and \(\; C(x_3, y_3) \),
then coordinates of midpoint of \(\; BC\;\) :
\[\left(\frac{x_2+x_3}{2}, \frac{y_2+y_3}{2} \, \right)\]
Now, if point \(\; O\) divides median \(\; AD\;\) in the ratio of \(\; 2:1\;\) then,
\[x_1= 4,\, y_1= 2,\, x_2= \frac{x_2+x_3}2,\, y_2= \frac{y_2+y_3}2\]
\[m=2, \; n=1\]
∴ Coordinates of the
centroid of the triangle
\[ \left(\,
\frac { 2 \cdot \frac{x_2+x_3}2 + 1 \cdot 4 } { 2 + 1},
\frac { 2 \cdot \frac{y_2+y_3}2 + 1 \cdot 2 } { 2 + 1} \,
\right)
\]
\[\left(\frac{x_2+x_3+4}{3},\, \frac{y_2+y_3+2}{3} \, \right)\]
Q. 8: ABCD is a rectangle formed by the points
\(A(-1, \;-1), B(-1, \;4), C(5, \;4)\) and \(D(5, \;-1)\).
P, Q, R and S are the mid-points of AB, BC, CD and DA respectively.
Is the quadrilateral PQRS a square? a rectangle? or a rhombus?
Justify your answer.
Solution
Step #1: Find midpoints P, Q, R, S.
Step #2: Find lengths of the sides PQ, QR, RS, SP.
Step #3: Find lengths of the diagonals PR, QS.
Step #4: Examine properties of quadrilateral PQRS
Step #1:
Find the coordinates of the midpoints P, Q, R, S with the
midpoint formula:
\(\; P \) is the midpoint of \(\; AB \;\) while
\[x_1= -1,\, y_1= -1,\, x_2= -1,\, y_2= 4\]
∴ Coordinates of point \(\;P\)
\[\left(\frac{(-1)+(-1)}{2}, \frac{(-1)+4}{2} \, \right)\]
\[ = \left( \frac{-2}2, \frac32 \right) \]
\[ = \left( -1, \frac32 \right) \]
\(\; Q \) is the midpoint of \(\; BC \;\) while
\[x_1= -1,\, y_1= 4,\, x_2= 5,\, y_2= 4\]
∴ Coordinates of point \(\;Q\)
\[\left(\frac{(-1)+5}{2}, \frac{4+4}{2} \, \right)\]
\[ = \left( \frac42, \frac82 \right) \]
\[ = \left( 2, 4 \right) \]
\(\; R \) is the midpoint of \(\; CD \;\) while
\[x_1= 5,\, y_1= 4,\, x_2= 5,\, y_2= -1\]
∴ Coordinates of point \(\;R\)
\[\left(\frac{5+5}{2}, \frac{4+(-1)}{2} \, \right)\]
\[ = \left( \frac{10}2, \frac32 \right) \]
\[ = \left( 5, \frac32 \right) \]
\(\; S \) is the midpoint of \(\; AD \;\) while
\[x_1= -1,\, y_1= -1,\, x_2= 5,\, y_2= -1\]
∴ Coordinates of point \(\;S\)
\[\left(\frac{(-1)+5}{2}, \frac{(-1)+(-1)}{2} \, \right)\]
\[ = \left( \frac42, \frac{-2}2 \right) \]
\[ = \left( 2, -1 \right) \]
Step #2:
Find lengths the of sides PQ, QR, RS, SP with the
distance formula:
Step #4:
Examine the properties of the quadrilateral.
We find that all the sides of the quadrilateral
PQRS are equal to each other: \(\;PQ = QR = RS = SP \).
So, it is either a square or a rhombus.
Since the diagonals of the quadrilateral PQRS are not equal, \(\;(PR \ne QS) \),
PQRS is a rhombus Ans.
