If -4- 0- and -1- -1- are two vertices of a triangle of area 4 sq- units- then its third vertex lies ony - x5x - y - 12 - 0x - 5y - 4 - 0x - 5y - 12 - 0

Question

Toppr Admin · Accepted Answer

Let the third vertex be -x-y- Then-12-x2223-x2223-x2223-x2223-xy1-x2212-4011-x2212-11-x2223-x2223-x2223-x2223-xB1-4-x21D2-x-5y-4-xB1-8-x21D2-x-5y-12-0orx-5y-x2212-4-0Hence- the third vertex lies on x-5y-12-0 or -xA0-x-5y-x2212-4-0

If (−4,0) and (1,−1) are two vertices of a triangle of area 4 sq. units, then its third vertex lies ony=x5x+y+12=0x+5y−4=0x+5y+12=0

Let the third vertex be (x,y). Then,12∣∣ ∣∣xy1−4011−11∣∣ ∣∣=±4⇒x+5y+4=±8⇒x+5y+12=0orx+5y−4=0Hence, the third vertex lies on x+5y+12=0 or x+5y−4=0

If $(- 4, 0)$ and $(1, - 1)$ are two vertices of a triangle of area $4$ sq. units, then its third vertex lies on
$y = x$
$5 x + y + 12 = 0$
$x + 5 y - 4 = 0$
$x + 5 y + 12 = 0$