Mathematics: I giving out 50 points

I giving out 50 points - id508708135, angellinelittle

Ответ:

memeE15

06.01.2023

(i) See attachment for further details, (ii) $y = -2\cdot x +14$ (for the segment BC), $y = -\frac{1}{3} \cdot x + 4$ (for the segment AC), (iii) $l_{AB} \approx 4.472$ , $l_{BC} \approx 4.473$ , A = 10.002 , (iv) $l_{\perp} \approx 3.162$

Step-by-step explanation:

i) The sketched triangle is presented in the figure attached below.

ii) Given that BC is perpendicular to AB and one point is known. It is required to know the slope of the line that contains AB and later, the equation of the line is finally found. The explicit form of the line that contains AB is:

$x - 2\cdot y + 8 = 0$

$2\cdot y = x+8$

$y = \frac{1}{2} \cdot x + 4$

The slope of the line containing BC can be found by the following equivalence:

$m_{BC} = - \frac{1}{m_{AB}}$

$m_{BC} = -\frac{1}{\frac{1}{2} }$

$m_{BC} = -2$

Now, the equation of the line is constructed hereafter:

$y - 2 = -2 \cdot (x-6)$

$y - 2 = -2\cdot x + 12$

$y = -2\cdot x +14$ (for the segment BC)

And given points A (0,4) and C (6,2), equation of the line containing AC is easily found. First, the slope of the function is:

$m_{AC} = \frac{2-4}{6-0}$

$m_{AC} = - \frac{1}{3}$

The y-intercept of the function corresponds to the y-component of point A. Then:

b = 4

Finally, the equation of the line containing AC is:

$y = -\frac{1}{3} \cdot x + 4$ (for the segment AC)

(iii) Length of the segment AC is determined by means of the Pythagorean Theorem:

$l_{AC} = \sqrt{(6-0)^{2}+(2-4)^{2}}$

$l_{AC} \approx 6.325$

Now, the location of point is determined by solving the following system of equations:

$y = \frac{1}{2} \cdot x + 4$ (for the segment AB)

$y = -2\cdot x +14$ (for the segment BC)

The value of x is found after algebraic handling:

$\frac{1}{2}\cdot x + 4 = -2 \cdot x +14$

$\frac{5}{2}\cdot x = 10$

x = 4

And the value of y is:

$y = -2\cdot (4) + 14$

y = 6

The location of point B is (4,6).

The length of the segment AB is:

$l_{AB} = \sqrt{(4-0)^{2}+(6-4)^{2}}$

$l_{AB} \approx 4.472$

And the length of BC is:

$l_{BC} = \sqrt{l_{AC}^{2}-l_{AB}^{2}}$

$l_{BC} = \sqrt{6.325^{2}-4.472^{2}}$

$l_{BC} \approx 4.473$

Lastly, the area of the triangle is:

$A = \frac{1}{2}\cdot l_{AB} \cdot l_{BC}$

$A = \frac{1}{2}\cdot (4.472)\cdot (4.473)$

A = 10.002

(iv) The perpendicular contains point B and crosses the segment AC, the slope of the line equation containing such line segment is found by using the following equivalence:

$m = -\frac{1}{m_{AC}}$