1 Marks Question

Question 11 Mark

Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).

Answer

Let (-1, 6) divides line segment joining the points (-3, 10) and(6, -8) in k:1.
Using Section formula, we get
$ - 1 = \frac{{( - 3) \times 1 + 6 \times k}}{{k + 1}}$ $4 \Rightarrow - k - 1 = ( - 3 + 6k)$
⇒ −7k = −2 ⇒ k= $\frac{2}{7}$
Therefore, the ratio is $\frac{2}{7}:1$ which is equivalent to 2:7.
Therefore, (-1, 6) divides line segment joining the points (-3, 10) and (6, -8) in 2:7.

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Question 21 Mark

Find the coordinates of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2 : 3.

Answer

Let the coordinates of the required point be (x, y). Then,
$x = \frac{{{m_1}{x_2} + {m_2}{x_1}}}{{{m_1} + {m_2}}}$
$= \frac{{(2)(4) + (3)( - 1)}}{{2 + 3}}$
$= \frac{{8 - 3}}{5} = \frac{5}{5} = 1$
$y = \frac{{{m_1}{y_2} + {m_2}{y_1}}}{{{m_1} + {m_2}}}$
$ = \frac{{(2)( - 3) + (3)(7)}}{{2 + 3}}$
$ = \frac{{ - 6 + 21}}{5} = \frac{{15}}{5} = 3$
Hence, the required point is (1, 3).

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Question 31 Mark

Find the distance between the pair of points (a, b), (-a, -b)

Answer

(a, b), (-a, -b)
Required distance
$= \sqrt {{{( - a - a)}^2} + {{( - b - b)}^2}}$
$= \sqrt {{{( - 2a)}^2} + {{( - 2b)}^2}}$
$= \sqrt {4{a^2} + 4{b^2}}$
$= \sqrt {4({a^2} + {b^2})}$
$= 2\sqrt {{a^2} + {b^2}}$

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Question 41 Mark

Find the distance between the pair of points (-5, 7), (-1, 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 $C\left(x_1, y_1\right)$ and $D\left(x_2, y_2\right)$ are the points of trisection of the line segment joining the given points i.e., $B C=C D=D A$
Let $B C=C D=D A=k$, Point $C$ divides $B C$ and $C A$ as: $B C=k C A=C D+D A=k+k=2 k$ Hence the ratio between $B C$ and $C A$ is: $\frac{B C}{C A}=\frac{k}{2 k}=\frac{1}{2}$
Therefore, point $C$ divides $B A$ internally in the ratio 1:2 then by section formula we have that if a point $P(x, y)$ divides two points $P\left(x_1, y_1\right)$ and $Q\left(x_2, y_2\right)$ in the ratio $m: n$ then, the point $(x, y)$ is given by $(x, y)=\left(\frac{ mx _2+ mx _1}{m+ n }, \frac{ my _2+ ny _1}{m+ n }\right)$
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 $B D$ and $D A$ as:DA $=k B D=B C+C D=k+k=2 k$
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 $B A$ 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 51 Mark

Find the distance between the pair of points: (2, 3), (4, 1)

Answer

Applying Distance Formula to find distance between points (2, 3) and (4,1), we get
d = $\sqrt {{{(4 - 2)}^2} + {{(1 - 3)}^2}} = \sqrt {{{(2)}^2} + {{( - 2)}^2}} = \sqrt {4 + 4} = \sqrt 8 = 2\sqrt 2 \;units$

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Question 61 Mark

Find the ratio in which the y-axis divides the line segment joining the points (5, -6) and (-1, -4). Also find the point of intersection.

Answer

Let the point on y-axis be P(0, y) and AP: PB = K: 1
Therefore $\frac { 5 - k } { k + 1 } = 0$ gives k = 5
Hence required ratio is 5: 1
$y = \frac { - 4 ( 5 ) - 6 } { 6 } = \frac { - 13 } { 3 }$
Hence point on y-axis is $\left( 0 , \frac { - 13 } { 3 } \right)$.

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Question 71 Mark

If the points A (6, 1), B (8, 2), C (9, 4) and D (p, 3) are the vertices of a parallelogram, taken in order, find the value of p.

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