[5 marks sum]

Question 15 Marks

The points $(2, -1), (-1, 4)$ and $(-2, 2)$ are mid points of the sides of a triangle. Find its vertices

Answer

Let $A(x_1, y_1), B(x_2,y_2)$ and $C (x_3,y_3)$ be the coordinates of the vertices of $\triangle ABC.$
Midpoint of $AB$, i.e.$D$
$D(2,-1)=D\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$
$2=\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}=-1$
$x_1+x_2=4......(1)$
$y_1+y_2=-2.....(2)$
Similarly
$x_1+x_2=-2 .....(3)$
$y_1+y_3=8.....(4)$
$x_1+x_3=-4.....(5)$
$y_2+y_3=4 .....(6)$
Adding (1), (3) and (5), we get,
$2\left(x_1+x_2+x_3\right)=-2$
$x_1+x_2+x_3=-1$
$4+x_3=-1(\text { from(1)) }$
$x_3=-5$
from (3)
$x_1-5=-2$
$x_1=3$
From (5)
$x_2-5=-4$
$x_2=1$
Adding (2), (4) and (6), we get,
$2\left(y_1+y_2+y_3\right)=10$
$y_1+y_2+y_3=5$
$-2+y_3=5($ from (2) $)$
$y_3=7$
From (4)
$y_1+7=8$
$y_1=1$
From (6)
$y_2+7=4$
$y_2=-3$
Thus, the co-ordinates of the vertices of $\triangle ABC$ are $(3,1), (1,-3)$ and $(-5,7)$

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Question 25 Marks

Given a line segment AB joining the points A $(-4, 6)$ and B $(8, -3).$ Find:
(i) the ratio in which AB is divided by the y-axis
(ii) find the coordinates of the point of intersection
(iii) the length of AB.

Answer

$(i)$ Let the required ratio be $m_1 : m_2$
Consider $A(-4, 6) = (x_1, y_1); B(8, −3) = (x_2 , y_2)$ and let
$P(x, y)$ be the point of intersection of the line segment
And the y-axis
By section formula, we have,
$x=\frac{m_1 x_2+m_2 x_1}{m_1+m_2}, y=\frac{m_1 y_2+m_2 y_1}{m_1+m_2} $
$\Rightarrow x=\frac{8 m_1-4 m_2}{m_1+m_2}, y=\frac{-3 m_1+6 m_2}{m_1+m_2}$
The equation of the y-axis is $x = 0$
$\Rightarrow x=\frac{8 m_1-4 m_2}{m_1+m_2}=0$
$ \Rightarrow 8 m_1-4 m_2=0$
$ \Rightarrow 8 m_1=4 m_2$
$ \Rightarrow \frac{m_1}{m_2}=\frac{4}{8} $
$\Rightarrow \frac{m_1}{m_2}=\frac{1}{2}$
$(ii)$ from the previous subpart, we have,
$\frac{m_1}{m_2}=\frac{1}{2}$
$\Rightarrow m_1=k$ and $m_2=2 k$, where $k$
Is any constant.
Also, we have,
$\Rightarrow x=\frac{8 m_1-4 m_2}{m_1+m_2}, y=\frac{-3 m_1+6 m_2}{m_1+m_2}$
$ \Rightarrow x=\frac{8 k-4 \times 2 k}{k+2 k}, y=\frac{-3 k+6 \times 2 k}{k+2 k} $
$ \Rightarrow x=\frac{8 k-8 k}{3 k}, y=\frac{-3 k+12 k}{3 k} $
$ \Rightarrow x=\frac{0}{3 k}, y=\frac{9 k}{3 k} $
$ \Rightarrow x=0, y=3$
Thus, the point of intersection is p $(0, 3)$
$(iii)$ The length of AB = distance between two points A and B.
The distance between two given points
$A\left(x_1, y_1\right)$ and $B \left(x_2, y_2\right)$ is given by,
$\text { Distance } AB =\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} $
$=\sqrt{(8+4)^2+(-3-6)^2} $
$ =\sqrt{12^2+9^2} $
$ =\sqrt{144+81} $
$ =\sqrt{225} $
$ =15 \text { units }$

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Mathematics [5 marks sum]

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