Question

Find the equation of straight line passing through the centroid of the triangle with vertices (2,6) (-4,3) (2,-3) and parallel to the given straight line 2x+5y=3.

Answer 1

Answer:

2x + 5y - 10 = 0

Step-by-step explanation:

Inorder to find the required equation line, we need slope and one point through which the line passes. Given that the line passes through the centroid of a traingle whose vertices are given.

Formula for centroid:

$\boxed{ \left(\mathrm{ \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} }\right)}$

Here,

• (x1, y1), (x2, y2) and (x3, y3) are vertices of triangle.

Let's assume that:

• (x1, y1) = (2, 6)
• (x2, y2) = (-4, 3)
• (x3, y3) = (2, -3)

The centroid is given by:

${ \implies \left(\mathrm{ \dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3} }\right)}$

${ \implies \left(\mathrm{ \dfrac{2 + ( - 4 ) + 2}{3}, \dfrac{6 + 3 + ( - 3)}{3} }\right)}$

${ \implies \left(\mathrm{ \dfrac{0}{3}, \dfrac{6}{3} }\right)}$

${ \implies \left(\mathrm{ 0, 2 }\right)}$

Therefore the line passes through (0, 2).

We know that slope of any two parallel lines is same.

So the slope of the required line will be same as the slope of line 2x + 5y = 3.

Slope of 2x + 5y = 3 is given by:

$\sf \longrightarrow m = - \dfrac{ coefficient \: of \: x}{coefficient \: of \: y}$

$\sf \longrightarrow m = - \dfrac{ 2}{5}$

Now the equation of line in point slope form is given by:

$\tt \implies y - y_1 = m(x - x_1)$

$\tt \implies y -2 = - \dfrac{ 2}{5} (x -0)$

$\tt \implies (y -2 )5= - 2x$

$\tt \implies 5y - 10= - 2x$

$\tt \implies 2x + 5y - 10=0$

This is the required equation.