Respuesta :
Answer:
[tex] A(\triangle ABC) = 7.2\: units^2 [/tex]
Step-by-step explanation:
C = (0, 7)
Since, point A lies in 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:
[tex] AC= \sqrt{(0-0)^2 +(7-1)^2} [/tex]
[tex] AC= \sqrt{(0)^2 +(6)^2 }[/tex]
[tex] AC= \sqrt{0 +36 }[/tex]
[tex] AC= \sqrt{36 }[/tex]
[tex] AC= 6\: units[/tex]
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:
[tex] CB =\frac{|2\times 0+(-1)\times 7+1|}{\sqrt {2^2 +(-1)^2} } [/tex]
[tex] CB =\frac{|0-7+1|}{\sqrt {4 +1} } [/tex]
[tex] CB =\frac{|-6|}{\sqrt {5} } [/tex]
[tex] CB =\frac{6}{2.24} [/tex]
[tex] CB =2.68\: units [/tex]
By Pythagoras Theorem:
[tex] AB=\sqrt{AC^2 - CB^2} [/tex]
[tex] AB=\sqrt{6^2 - (2.68)^2} [/tex]
[tex] AB=\sqrt{36 - 7.18} [/tex]
[tex] AB=\sqrt{28.82} [/tex]
[tex] AB=5.36842621\: units [/tex]
[tex] AB\approx 5.37\: units [/tex]
[tex] A(\triangle ABC) = \frac{1}{2} \times AB\times CB[/tex]
[tex] A(\triangle ABC) = \frac{1}{2} \times 5.37\times 2.68[/tex]
[tex] A(\triangle ABC) = \frac{1}{2} \times 14.3916[/tex]
[tex] A(\triangle ABC) = 7.1958[/tex]
[tex] A(\triangle ABC) = 7.2\: units^2 [/tex]