Question

EXPLANATION NEEDED
1) Find the slope, x-intercept and y-intercept of line 4x - 6y - 24 = 0.
{answer should come $\frac{2}{3}, 6, -4$}

2) Reduce the equation 3x - 2y + 6 = 0 to the double-intercept form and find the x-intercept and y-intercept. {answer should come $\frac{x}{-2} +\frac{y}{3}=1, (-2), 3$}

3) Th centroid of a triangle is (1,4) and co-ordinates of two of its vertices are (4,-3) and (-9,7). Find the area of the triangle. (answer should come 183/2)

Answer 1

4x - 6 y - 24 =0
divide by  the constant term and write it so that RHS has value 1.
   x / 6 - y / 4 = 1 
x intercept is 6  and  y intercept is 4.

write y term on LHS and x term and constant on the RHS.
   6 y = 4  x  - 24        =>       y =  2/3  x  - 4
     slope is  2/3...          OR, slope =  - (coeff of x ) / (coeff of y)
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3 x - 2y + 6 = 0
     =>   y / 3 - x / 2 = 1
       =>  x intercept = - 2      and    y intercept = 3
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x coordinate of centroid = 1 = average of x - coordinates of three vertices
y coordinate of centroid = 4 = average of y - coordinates of three vertices

Let the coordinates of the third vertex C be = (x3,y3)
         1 = (4-9+x3) / 3        => x = 8
         4 = (-3+7+y3) / 3  => y = 0

use the formula for the area of triangle in terms of coordinates of the vertices.

  area = | x1 (y2 - y3) + x2 (y3 - y1) + x3 (y1 - y2) | / 2
         = | 4 (7 - 0) - 9 ( 0 +3 ) + 8 ( -3 - 7 ) | / 2
       = 79/2

OR,  by finding the lengths of sides as,
      AC² = (8-4)² + (0+3)² = 5²
      BC² = (8+9)²+(0+7)² = 338 = 13² * 2
      AB² = (9+4)²+(7+3)²  = 269
s = semi perimeter = (5 + 13√2 + √269) / 2
the area of the triangle could be found using the formula √[ s(s-a)(s-b)(s-c) ]