5 Marks Questions

Question 15 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 $P (x_1, y_1) Q(x_2, y_2)$ and $R(x_3, y_3)$ 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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Question 25 Marks

To conduct Sports Day activities, in your rectangular-shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1m each. 100 flower pots have been placed at a distance of 1m from each other along AD, as shown in Figure. Niharika runs $\frac{1}{4}$th the distance AD on the 2nd line and posts a green flag. Preet runs $\frac{1}{5}$th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?

Answer

It can be observed that Niharika posted the green flag at $\frac{1}{4}$th of the distance AD i.e., $\frac{1}{4} \times 100 = 25m$ from the starting point of $2^{nd}$ line. Therefore, the coordinates of this point G is (2, 25)
Similarly, Preet posted a red flag at $\frac{1}{5}$th of the distance AD i.e., $\frac{1}{5} \times 100 = 20m$ from the starting point of $8^{th}$ line. Therefore, the coordinates of this point R are (8, 20)
Now we have the positions of posts by Preet and Niharika
According to distance formula, the distance between points $A(x_1, y_1)$ and $B(x_2, y_2)$ is given by
$D=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$
Distance between these flags by using distance formula,
$D=\sqrt{\left(8-2_{}\right)^{2}+\left(25_{}-{20}\right)^{2}}$
$=\sqrt{36+25} \mathrm {m}$
$=\sqrt{61} \mathrm{m}$
The point at which Rashmi should post her blue flag is the mid-point of the line joining these points. Let this point be A (X,Y)
Now by midpoint formula,
$(X, Y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$
$X=\left(\frac{2+8}{2}\right)=5$
$Y=\left(\frac{25+20}{2}\right)=22.5$
Hence, A (X,Y) = (5, 22.5)
Therefore, Rashmi should post her blue flag at 22.5m on the $5^{th}$ line.

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Question 35 Marks

Find the coordinates of the points of trisection of the line segment joining $(4, –1)$ and $(–2, –3).$

Answer

The points of trisection means that the points which divide the line into three equal parts. From the figure, it is clear that C, and D are these two points. Let $C (x_1, y_1)$ and $D (x_2, y_2)$ are the points of trisection of the line segment joining the given points i.e., $BC = CD = DA$
Let $BC = CD = DA = k$, Point C divides BC and CA as: $BC = kCA = CD + DA = k + k = 2k$
Hence the ratio between BC and CA is: $\frac{B C}{C A}=\frac{k}{2 k}=\frac{1}{2}$
Therefore, point C divides BA internally in the ratio 1:2 then by section formula we have that if a point P(x, y) divides two points $P (x_1, y_1)$ and $Q (x_2, y_2)$ in the ratio m:n then, the point (x, y) is given by $(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{mx}_{2}+\mathrm{nx}_{1}}{\mathrm{m}+\mathrm{n}}, \frac{\mathrm{my}_{2}+\mathrm{ny}_{1}}{\mathrm{m}+\mathrm{n}}\right)$
Therefore C(x, y) divides B(–2, –3) and A(4,–1) in the ratio 1:2, then
$C(x, y)=\left(\frac{(1 \times 4)+(2 \times-2)}{1+2}, \frac{(1 \times-1)+(2 \times-3)}{1+2}\right)$
$C(x, y)=\left(\frac{4-4}{1+2}, \frac{-1-6}{1+2}\right)$
$C(x, y)=\left(0, \frac{-7}{3}\right)$
Point D divides the BD and DA as:DA = kBD = BC + CD = k + k = 2k
Hence the ratio between BD and DA is: $\frac{B D}{D A}=\frac{2 k}{k}=\frac{2}{1}$
The point D divides the line BA in the ratio 2:1
So now applying section formula again we get,
$D(x, y)=\left(\frac{(2 \times 4)+(1 \times-2)}{2+1}, \frac{(2 \times-1)+(1 \times-3)}{2+1}\right)$
$D(x, y)=\left(\frac{8-2}{3}, \frac{-2-3}{3}\right)$
$D(x, y)=\left(\frac{6}{3}, \frac{-5}{3}\right)$
$D(x, y)=\left(2, \frac{-5}{3}\right)$

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Question 45 Marks

In a classroom, 4 friends are seated at the four points A, B, C and D as shown in Fig. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, Don’t you think ABCD is a square? Chameli disagrees. Using distance formula, find which of them is correct.

Answer

It can be seen that A (3, 4), B (6, 7), C (9, 4), and D (6, 1) are the positions of 4 friends

Distance between two points $A(x_{1, }y_1)$ and $B(x_2, y_2)$ is given by
$D=\sqrt{\left( x_{2} - x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$
Hence,
$AB = [(3- 6)^2 + (4 - 7)^2]^{1/2}$
= $\sqrt{9+9}=\sqrt{18}$
= $3 \sqrt{2}$
$BC = [(6 - 9)^2 + (7- 4)^2]^{1/2}$
= $\sqrt{9+9}=\sqrt{18}$
= $3 \sqrt{2}$
$CD = [(9- 6)^2 + (4- 1)^2]^{1/2}$
= $\sqrt{9+9}=\sqrt{18}$
= $3 \sqrt{2}$
$AD = [(3 - 6)^2 + (4 - 1)^2]^{1/2}$
= $\sqrt{9+9}=\sqrt{18}$
= $3 \sqrt{2}$
Diagonal $AC = [(3 - 9)^2 + (4 - 4)^2]^{1/2}$
= $\sqrt{36+0}$ = 6
Diagonal $BD = [(6- 6)^2 + (7- 1)^2]^{1/2}$
= $\sqrt{36+0}$ = 6
It can be seen that all sides of quadrilateral ABCD are of the same length and diagonals are of the same length
Therefore, ABCD is a square and hence, Champa was correct.

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