Question

Hi:) I need help with part 9(ii) ? Thanks :)!

Answer 1

Answer:

see explanation

Step-by-step explanation:

Given

2BD = 7BT , then

2(d - b ) = 7(t - b) ← distribute both sides

2d - 2b = 7t - 7b ( add 7b to both sides )

2d + 5b = 7t, thus

2$\left[\begin{array}{ccc}10.5\\10\\\end{array}\right]$ + 5$\left[\begin{array}{ccc}0\\3\\\end{array}\right]$ = 7t

$\left[\begin{array}{ccc}21\\20\\\end{array}\right]$ + $\left[\begin{array}{ccc}0\\15\\\end{array}\right]$ = 7t

$\left[\begin{array}{ccc}21\\35\\\end{array}\right]$ = 7t , thus

t = $\frac{1}{7}$$\left[\begin{array}{ccc}21\\35\\\end{array}\right]$ = $\left[\begin{array}{ccc}3\\5\\\end{array}\right]$

Hence T = (3, 5 )

Answer 2

Answer:

Please see the attached pictures for the full solution.

We reject x= -3 since x has to be between x=0 and x=10.5 since T lies on line BD.

We reject y= 1 since the y value has to be between 3 and 10 for T to lie on line BD.

Useful formula:

gradient=(y1-y2)/(x1-x2)

Further explanation:

1st picture, from line 12 to 13: cross multiply

line 14: expand

line 15: bring 9 over (or +9 on both sides)

line 16: ÷3 throughout

last picture: we have to show '(reject)' because whenever we introduce a square root, there must be a '±' sign.