Question

PLEASE HELP I AM SO STUCK. IT IS DUE ~TOMORROW.

Answer 1

Answer:

$A(\triangle ABC) = 7.2\: units^2$

Step-by-step explanation:

C = (0, 7)

Since, point A lies on y-axis, so it's x coordinate will be zero. Line y = 2x + 1 passes through point A.

Therefore, plug x = 0 in y = 2x +1, we find:

y = 2*0 + 1 = 0 + 1 = 1

Hence, coordinates of point A are (0, 1).

Now, by distance formula:

$AC= \sqrt{(0-0)^2 +(7-1)^2}$

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

$AC= \sqrt{0 +36 }$

$AC= \sqrt{36 }$

$AC= 6\: units$

Length of perpendicular CB dropped from point C (0, 7) to line y = 2x + 1 or 2x - y + 1 = 0 can be obtained as given below:

$CB =\frac{|2\times 0+(-1)\times 7+1|}{\sqrt {2^2 +(-1)^2} }$

$CB =\frac{|0-7+1|}{\sqrt {4 +1} }$

$CB =\frac{|-6|}{\sqrt {5} }$

$CB =\frac{6}{2.24}$

$CB =2.68\: units$

By Pythagoras Theorem:

$AB=\sqrt{AC^2 - CB^2}$

$AB=\sqrt{6^2 - (2.68)^2}$

$AB=\sqrt{36 - 7.18}$

$AB=\sqrt{28.82}$

$AB=5.36842621\: units$

$AB\approx 5.37\: units$

$A(\triangle ABC) = \frac{1}{2} \times AB\times CB$

$A(\triangle ABC) = \frac{1}{2} \times 5.37\times 2.68$

$A(\triangle ABC) = \frac{1}{2} \times 14.3916$

$A(\triangle ABC) = 7.1958$

$A(\triangle ABC) = 7.2\: units^2$