Question

pls answer this question

Answer 1

(a)

(i)

$\displaystyle\longrightarrow\sf {|\vec {AB}|=\sqrt {(-8)^2+15^2}}$

$\displaystyle\longrightarrow\sf {|\vec {AB}|=\sqrt {64+225}}$

$\displaystyle\longrightarrow\sf {|\vec {AB}|=\sqrt {289}}$

$\displaystyle\longrightarrow\sf {\underline {\underline {|\vec {AB}|=17}}}$

(ii)

Let the coordinates of B be $\displaystyle\sf {(x_b,\ y_b).}$

Then,

$\displaystyle\begin {aligned}\longrightarrow\quad&\sf {x_b-10=-8,}&\quad&\sf{y_b-1=15}\\\\\longrightarrow\quad&\sf {x_b = -8 + 10,}&\quad&\sf{y_b=15 + 1}\\\\\longrightarrow\quad&\sf {x_b=2,}&\quad&\sf{y_b=16}\end{aligned}$

Hence coordinates of B is $\displaystyle\sf {\underline {\underline {(2,\ 16)}}.}$

(b)

Since $\displaystyle\sf {\vec {CD}=3\vec {AB},}$

$\displaystyle\longrightarrow\sf {\vec {CD}=\binom{-24}{45}}$

(i)

Let the coordinates of D be $\displaystyle\sf {(x_d,\ y_d).}$

Then,

$\displaystyle\begin {aligned}\longrightarrow\quad&\sf {x_d-42=-24,}&\quad&\sf{y_d-16=45}\\\\\longrightarrow\quad&\sf {x_d = -24+42,}&\quad&\sf{y_d=45+16}\\\\\longrightarrow\quad&\sf {x_d=18,}&\quad&\sf{y_d=61}\end{aligned}$

Hence coordinates of D is $\displaystyle\sf {\underline {\underline {(18,\ 61)}}.}$

(ii)

$\displaystyle\longrightarrow\sf {\vec {AD}=\binom{18-10}{61-1}}$

$\displaystyle\longrightarrow\sf {\underline {\underline {\vec {AD}=\binom{8}{60}}}}$

(c)

(i)

$\displaystyle\longrightarrow\sf {\vec {AE}=\binom{k-10}{16-1}}$

$\displaystyle\longrightarrow\sf {\underline {\underline {\vec {AE}=\binom{k-10}{15}}}}$

(ii)

Since $\displaystyle\sf {AED}$ is a straight line and $\displaystyle\sf {\vec {CD}=3\vec {AB},}$

$\displaystyle\longrightarrow\sf {\vec {ED}=3\vec {AE}}$

Because the triangles $\displaystyle\sf {ABE}$ and $\displaystyle\sf {CDE}$ are similar and they're scaled by a factor 3.

So,

$\displaystyle\longrightarrow\sf {\binom {18-k}{61-16}=3\binom {k-10}{15}}$

$\displaystyle\longrightarrow\sf {\binom {18-k}{45}=\binom {3k-30}{45}}$

From this,

$\displaystyle\longrightarrow\sf {18-k=3k-30}$

$\displaystyle\longrightarrow\sf {\underline {\underline {k=12}}}$

(d)

As said earlier, the triangles $\displaystyle\sf {ABE}$ and $\displaystyle\sf {CDE}$ are similar and they're scaled by a factor 3.

Hence their areas are scaled by a factor of 3² = 9.

$\displaystyle\therefore\sf {\underline {\underline {\dfrac {Area\ of\ triangle\ ABE}{Area\ of\ triangle\ CDE}=\dfrac {1}{9}}}}$