Question

The point a(0,0) , b(1, 7) , c(5,1) are the vertices of a triangle . find the length of the perpendicular drawn from a to bc and hence find the area of triangle abc

Answer 1

Answer:

it can be the point of c and are equal to each other and it is the BODMAS rule of the triangle

Step-by-step explanation:

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Answer 2

Answer:

Length of the perpendicular is 4.71 units and area of the triangle ABC is 17 square units.

Step-by-step explanation:

Length (d) of the perpendicular drawn from a point (h, k) to the line ax + by + c is represented by,

d = $\frac{|ah+bk+c|}{\sqrt{a^{2}+b^{2}}}$

We will find the equation of the line BC passing through two points b(1, 7) and c(5, 1)

Slope (m) of the line BC = $\frac{y-y'}{x-x'}=\frac{(7-1)}{1-5}$

m = $-\frac{6}{4}$

m = $-\frac{3}{2}$

Equation of the line passing through (1, 7) and slope $(-\frac{3}{2})$ will be

$y-7=-\frac{3}{2}(x-1)$

$2y-14=-3x+3$

3x + 2y - 17 = 0

Now length of perpendicular from a(0, 0) to the given line

d = $\frac{|(-17)|}{\sqrt{3^{2}+2^{2}}}$

d = $\frac{17}{\sqrt{13}}$

d = 4.71 units

Length of segment AB = $\sqrt{(1-5)^{2}+(7-1)^{2}}=\sqrt{16+36}$

AB = $\sqrt{52}$

AB = $2\sqrt{13}$ units

Area of the given triangle ABC = $\frac{1}{2}(AB)(d)$

Area = $\frac{1}{2}(2\sqrt{13})(\frac{17}{\sqrt{13}})$

Area = 17 square units.

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