Question

If x1 x2 x3 are in ap with common difference d1 and y1 y2 y3 are in ap with common difference d2 then area of triangle formed by

Answer 1

Given:
———-

x2 - x1 = x3 - x2 = d1

x1 - x2 = x2 - x3 = -d1 —-> 1

y2 - y1 = y3 - y2 = d2

y1 - y2 = y2 - y3 = -d2 —-> 2


Area of the triangle = (1/2)*|(x1)*(y2 - y3) + (x2)*(y3 - y1) + (x3)*(y1 - y2)|

= (1/2)*|(x1)*(-d2) + (x2)*(y3 - y1) + (x3)*(-d2)|

= (1/2)*|(x1)*(-d2) + (x3)*(-d2) + (x2)*(y3 - y1) |

= (1/2)*|(-d2)*(x1 + x3) + (x2)*(y3 - y1)|

= (1/2)*|(-d2)*(x1 + x3) + (x2)*(d2 + y2 - y1)|

= (1/2)*|(-d2)*(x1 + x3) + (x2)*(d2 + d2)|

= (1/2)*|(-d2)*(x1 + x3) + (x2)*(2d2)|

= (1/2)*(d2)*|2x2 - x1 - x3|

= (1/2)*(d2)*|x2 + x2 - x1 - x3|

= (1/2)*(d2)*|(x2 - x1) + (x2 - x3)|

= (1/2)*(d2)*|d1 - d1|

= 1/2)*(d2)*0

= 0 ——> Answer