3 Marks Question

Question 13 Marks

Find the point on the $x-$axis which is equidistant from $(2, -5)$ and $(-2, 9).$

Answer

We know that a point on the $x-$axis is of the form $(x, 0)$. So, let the point $P(x, 0)$ be equidistant from $A(2, –5)$ and $B(–2, 9).$ Then
$PA = PB$
$ \Rightarrow P A^2=P B^2 $
$ \Rightarrow(2-x)^2+(-5-0)^2=(-2-x)^2+(9-0)^2$
$\Rightarrow 4 + x^2- 4x + 25 = 4 + x^2+ 4x + 81$
$\Rightarrow- 4x + 25 = 4x + 81$
$\Rightarrow 8x = -56$
$\Rightarrow \;x = \frac{{ - 56}}{8} = - 7$
Hence, the required point is $(-7, 0)$
Check:
$PA = \sqrt {{{\{ 2 - ( - 7)\} }^2} + {{( - 5 - 0)}^2}}$
$= \sqrt {81 + 25} = \sqrt {106}$
$PB = \sqrt {{{\{ - 2 - ( - 7)\} }^2} + {{(9 - 0)}^2}}$
$= \sqrt {25 - 81} = \sqrt {106}$
$\because PA = PB$
$\therefore$ Our solution is checked.

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Question 23 Marks

Name the type of quadrilateral formed, if any, by the points $(4, 5), (7, 6), (4, 3), (1, 2),$ and give a reason for your answer.

Answer

$(4, 5), (7, 6), (4, 3), (1, 2)$
Let $A \rightarrow (4, 5), B \rightarrow (7, 6), C \rightarrow (4, 3)$ and $D \rightarrow (1, 2)$
Then, $AB = \sqrt {{{(7 - 4)}^2} + {{(6 - 5)}^2}}$
$= \sqrt {{{(3)}^2} + {{(1)}^2}} = \sqrt {9 + 1} = \sqrt {10}$
$BC = \sqrt {{{(4 - 7)}^2} + {{(3 - 6)}^2}}$
$\sqrt {{{( - 3)}^2} + {{( - 3)}^2}} = \sqrt {9 + 9} = \sqrt {18} = 3\sqrt 2$
$CD = \sqrt {{{(1 - 4)}^2} + {{(2 - 3)}^2}}$
$= \sqrt {{{( - 3)}^2} + {{( - 1)}^2}} = \sqrt {9 + 1} = \sqrt {10}$
$DA = \sqrt {{{(4 - 1)}^2} + {{(5 - 2)}^2}}$
$= \sqrt {9 + 9} = \sqrt {18} = 3\sqrt 2$
$AC = \sqrt {{{(4 - 4)}^2} + {{(3 - 5)}^2}} = 2$
$BD = \sqrt {{{(1 - 7)}^2} + {{(2 - 6)}^2}}$
$= \sqrt {36 + 16} = \sqrt {52}$
We see that
$AB = CD,$ opposite sides are equal
$BC = DA$
and $AC \ne BD .....$ Diagonals are unequal
Hence, the quadrilateral $ABCD$ is a parallelogram.

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Question 33 Marks

Name the type of quadrilateral formed, if any, by the points $(-3, 5), (3, 1), (0, 3), (-1, -4),$ and give a reason for your answer.

Answer

$(-3, 5), (3, 1), (0, 3), (-1, -4)$
Let A $\rightarrow (-3, 5), B \rightarrow (3, 1), C \rightarrow (0, 3)$ and $D \rightarrow (-1, -4)$
Then, $AB = \sqrt {{{(3 - ( - 3))}^2} + {{(1 - 5)}^2}} = \sqrt {{{(6)}^2} + {{( - 4)}^2}}$
$= \sqrt {36 + 16} = \sqrt {52} = 2\sqrt {13}$
$BC = \sqrt {{{(0 - 3)}^2} + {{(3 - 1)}^2}} = \sqrt {9 + 4} = \sqrt {13}$
$CD = \sqrt {{{( - 1 - 0)}^2} + {{( - 4 - 3)}^2}}$$= \sqrt {1 + 49} = \sqrt {50}$
$DA = \sqrt {[( - 3) - {{( - 1)}]^2} + {{[5 - ( - 4)]}^2}}$$= \sqrt {4 + 81} = \sqrt {85}$
$AC = \sqrt {{{[0 - ( - 3)]}^2} + {{(3 - 5)}^2}} = \sqrt {13}$
$BD = \sqrt {{{( - 1 - 3)}^2} + {{( - 4 - 1)}^2}} = \sqrt {41}$
We see that $BC + AC = AB$
Hence, the points $A, B$ and $C$ are collinear.
So, $ABCD$ is not a quadrilateral.

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Question 43 Marks

Name the type of quadrilateral formed, if any, by the points $(-1, -2), (1, 0), (-1, 2), (-3, 0),$ and give a reason for your answer.

Answer

$(-1, -2), (1, 0), (-1, 2), (-3, 0)$
Let $A \rightarrow (-1, -2), B \rightarrow (1, 0)$
$C \rightarrow (-1, 2)$ and $D \rightarrow (-3, 0)$
Then,
$AB = \sqrt {[1 - {{( - 1)}]^2} + {{[0 - ( - 2)]}^2}}$
$= \sqrt {{{(2)}^2} + {{(2)}^2}} = \sqrt {4 + 4} = \sqrt 8 = 2\sqrt 2$
$BC = \sqrt {{{( - 1 - 1)}^2} + {{(2 - 0)}^2}}$
$ = \sqrt {{{( - 2)}^2} + {{(2)}^2}} = \sqrt {4 + 4} = \sqrt 8 = 2\sqrt 2$
$CD = \sqrt {[ (- 3) - {{( - 1)}]^2} + {{(0 - 2)}^2}}$
$= \sqrt {{{( - 2)}^2} + {{( - 2)}^2}} = \sqrt {4 + 4} = \sqrt 8 = 2\sqrt 2$
$DA = \sqrt {{{[( - 1) - ( - 3)]}^2} + {{( - 2 - 0)}^2}}$
$= \sqrt {{{( 2)}^2} + {{( - 2)}^2}} = \sqrt {4 + 4} = \sqrt 8 = 2\sqrt 2$
$AC = \sqrt {[( - 1) - {{( - 1)}]^2} + {{[( 2) - ( - 2)]}^2}} = 4$
$BD = \sqrt {[( - 3) - {{( 1)}]^2} + {{(0 - 0)}^2}} = 4$
Since $AB = BC = CD = DA ($i.e., all the four sides of the quadrilateral $ABCD$ are equal$)$ and $AC = BD ($i.e. diagonals of the quadrilateral $ABCD$ are equal$).$ Therefore, $ABCD$ is a square.

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Question 53 Marks

Find the distance between the points $(0, 0)$ and $(36, 15).$ Can you now find the distance between the two towns $A$ and $B$ discussed in Section $7.2.$

Answer

Distance between two points
Given: Points $A(0, 0), B(36, 15)$

For two points $\mathrm{A}\left(\mathrm{x}_1, \mathrm{y}_1\right)$ and $\mathrm{B}\left(\mathrm{x}_2, \mathrm{y}_2\right)$,
Distance is given by $f = \sqrt{{({\mathrm{{x_2-x_1}}})^2}+{({\mathrm{{y_2-y_1}}})^2}}$ Distance between $(0, 0)$ and $(36, 15)$ is:
$f = \sqrt{{({\mathrm{{36-0}}})^2}+{({\mathrm{{15-0}}})^2}}$
$f = \sqrt{1296+225}$
$f = \sqrt{1521}= 39$
Hence Distance between points $A$ and $B$ is $39$ units
Yes, we can find the distance between the given towns $A$ and $B.$ Let us take town $A$ at origin point $(0, 0)$
Hence, town $B$ will be at point $(36, 15)$ with respect to town $A$
And, as calculated above, the distance between town $A$ and $B$ will be $39\ km$

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Question 63 Marks

Find the coordinates of the point which divides the line segment joining the points $(4, -3)$ and $(8, 5)$ in the ratio $3 : 1$ internally.

Answer

Let coordinates of the required point be $R(x, y).$ This means $R$ divides the join of $P(4, -3)$ and $Q(8, 5)$ in the ratio $3:1$ internally.
Using the Section formula for internal division, here $\mathrm{x}_1=4, \mathrm{y}_1=-3, \mathrm{x}_2=8, \mathrm{y}_2=5, \mathrm{~m}=3, \mathrm{n}=1$
$\Rightarrow(x,y) = \left( \frac { mx _ { 2 } +n x _ { 1 } } { m+n} , \frac { m y _ { 2 } + ny _ { 1 } } { m+n } \right)$
$\Rightarrow(x,y) = (\frac { 3 ( 8 ) + 1 ( 4 ) } { 3 + 1 },\frac { 3 ( 5 ) + 1 ( - 3 ) } { 3 + 1 }) \Rightarrow (x,y) = ( \frac { 24 + 4 } { 4 } , \frac { 15-3} { 4 })$
$\Rightarrow (x,y) = ( \frac { 28 } { 4 } , \frac { 12} { 4 }) =(7, 3) \Rightarrow x= 7$ and $y= 3$
Thus, the coordinates of $R (x,y) = (7,3)$

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Question 73 Marks

Show that the points $A (1, 7), B (4, 2), C (-1, -1)$ and $D (-4, 4)$ are the vertices of a square.

Answer

Let $A(1, 7), B(4, 2), C(-1, -1)$ and $D(-4, 4)$ be the given points. One way of showing that $ABCD$ is a square is to use the property that all its sides should be equal both its diagonals should also be equal. Now,
$A B = \sqrt { ( 1 - 4 ) ^ { 2 } + ( 7 - 2 ) ^ { 2 } } = \sqrt { 9 + 25 } = \sqrt { 34 }$
$B C = \sqrt { ( 4 + 1 ) ^ { 2 } + ( 2 + 1 ) ^ { 2 } } = \sqrt { 25 + 9 } = \sqrt { 34 }$
$C D = \sqrt { ( - 1 + 4 ) ^ { 2 } + ( - 1 - 4 ) ^ { 2 } } = \sqrt { 9 + 25 } = \sqrt { 34 }$
$D A = \sqrt { ( 1 + 4 ) ^ { 2 } + ( 7 - 4 ) ^ { 2 } } = \sqrt { 25 + 9 } = \sqrt { 34 }$
$A C = \sqrt { ( 1 + 1 ) ^ { 2 } + ( 7 + 1 ) ^ { 2 } } = \sqrt { 4 + 64 } = \sqrt { 68 }$
$B D = \sqrt { ( 4 + 4 ) ^ { 2 } + ( 2 - 4 ) ^ { 2 } } = \sqrt { 64 + 4 } = \sqrt { 68 }$
Since, $AB = BC = CD = DA$ and $AC = BD,$ all the four sides of the quadrilateral $ABCD$ are equal and its diagonals $AC$ and $BD$ are also equal. Therefore, $ABCD$ is a square.

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Question 83 Marks

Do the points $(3, 2), (–2, –3)$ and $(2, 3)$ form a triangle$?$ If so, name the type of triangle formed.

Answer

Let us apply the distance formula to find the distances $PQ, QR$ and $PR,$ where
$P \leftrightarrow (3, 2)$,
$Q \leftrightarrow (–2, –3)$ and
$R \leftrightarrow (2, 3)$
are the given points. We have
$\mathrm{PQ}=\sqrt{(3+2)^{2}+(2+3)^{2}}=\sqrt{5^{2}+5^{2}}=\sqrt{50}=7.07(\text { approx. })$
$\mathrm{QR}=\sqrt{(-2-2)^{2}+(-3-3)^{2}}=\sqrt{(-4)^{2}+(-6)^{2}}=\sqrt{52}=7.21(\text { approx. })$
$\mathrm{PR}=\sqrt{(3-2)^{2}+(2-3)^{2}}=\sqrt{1^{2}+(-1)^{2}}=\sqrt{2}=1.41(\text { approx. })$
Since the sum of any two of these distances is greater than the third distance, therefore, the points $P, Q$ and $R$ form a triangle.
Also, $P Q^2+P R^2=Q R^2$, by the converse of Pythagoras theorem, we have $\angle P = 90^\circ$. Therefore, $PQR$ is a right triangle.

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Question 93 Marks

Check whether (5, –2), (6, 4) and (7, –2) are the vertices of an isosceles triangle.

Answer

Let A $\rightarrow$ (5, -2), B $\rightarrow$ (6, 4) and C $\rightarrow$ (7, -2)
Then,
$AB = \sqrt {{{(6 - 5)}^2} + {{(4 - ( - 2))}^2}} = \sqrt {{{(1)}^2} + {{(6)}^2}}$
$ = \sqrt {1 + 36} = \sqrt {37}$
$BC = \sqrt {{{(7 - 6)}^2} + {{( - 2 - 4)}^2}} = \sqrt {{{(1)}^2} + {{( - 6)}^2}}$
$= \sqrt {1 + 36} = \sqrt {37}$
CA = $\sqrt {{{(7 - 5)}^2} + {{( - 2 - (-2))}^2}} = \sqrt {{{(2)}^2} + {{(0)}^2}}$
We see that AB = BC $\neq$ CA
So, the A, B and C are vertices of an isosceles triangle.

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Question 103 Marks

Find the coordinates of the points of trisections of the line segment joining the points A(2, - 2) and B(- 7, 4)

Answer

Let $P$ and $Q$ be the points of trisection of $A B$.
ie., $A P=P Q=Q B$
Therefore, $P$ divides $A B$ internally in the ratio $1: 2$.
By applying the section formula coordinates of $P$ are:
$
\left(\frac{1(-7)+(2)(2)}{1+2}, \frac{1(4)+(2)(-2)}{1+2}\right)=(-1,0)
$
Now, $Q$ also divides $A B$ internally in the ratio $2: 1$.
So, the coordinates of $Q$ are: $\left(\frac{2(-7)+(1)(2)}{2+1}, \frac{(2)(4)+1(-2)}{2+1}\right)=(-4,2)$
Therefore, the coordinates of the points of trisection of the line segment
joining $A$ and $B$ are $(-1,0)$ and $(-4,2)$.

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Question 113 Marks

Find the coordinates of the points which divide the line segment joining A(- 2, 2) and B(2, 8) into four equal parts.

Answer

Let P (x₁, y₁) Q(x₂, y₂) and R(x₃, y₃) be the points which divide the line segment AB into four equal parts.

Then, P divides AB in the ratio 1 : 3 internally.
$x=\frac{mx_2+nx_1}{m+n}$
$\therefore x _ { 1 } = \frac { ( 1 ) ( 2 ) + ( 3 ) ( - 2 ) } { 1 + 3 }$
$= \frac { 2 - 6 } { 4 } = - \frac { 4 } { 4 } = - 1$
$y=\frac{my_2+ny_1}{m+n}$
$y _ { 1 } = \frac { ( 1 ) ( 8 ) + ( 3 ) ( 2 ) } { 1 + 3 }$
$= \frac { 8 + 6 } { 4 } = \frac { 14 } { 4 } = \frac { 7 } { 2 }$
So, $\mathrm { P } \rightarrow \left( - 1 , \frac { 7 } { 2 } \right)$
Also, Q divides AB in the ratio 1 : 1 i.e.
Q is the mid point of AB
$x _ { 2 } = \frac { - 2 + 2 } { 2 } = 0$
$y _ { 2 } = \frac { 2 + 8 } { 2 } = \frac { 10 } { 2 } = 5$
So, $Q \rightarrow ( 0,5 )$
and, R divides AB in the ratio 3 : 1
$\therefore x _ { 2 } = \frac { ( 3 ) ( 2 ) + ( 1 ) ( - 2 ) } { 3 + 1 }$
$= \frac { 6 - 2 } { 4 } = \frac { 4 } { 4 } = 1$
$y _ { 3 } = \frac { ( 3 ) ( 8 ) + ( 1 ) ( 2 ) } { 3 + 1 }$
$= \frac { 24 + 2 } { 4 } = \frac { 26 } { 4 } = \frac { 13 } { 2 }$
So, $\mathrm { R } \rightarrow \left( 1 , \frac { 13 } { 2 } \right)$

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