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if the base of the isosceles triangle joints the points (2,-4), (1,-3); the area is 9/2 . find the third vertex of the triangle.

please help me guys, show your solution

Note by Jeriel Villa
2 years, 8 months ago

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Jeriel, Let the coordinates be (x,y) using the distance formula, root{(2-x)^2 + (-4-y)^2}= root(1-x)^2+(-3-y)^2 [DISTANCE IS EQUAL AS ITS AN ISOSCELES TRI] = (2-x)^2 + (-4-y)^2 = (1-x)^2+(-3-y)^2 ["^2" MEANS SQUARE] = 4-4x+x^2+16-(-8y)+y^2= 1-2x+x^2+9-(-6y)+y^2 (USING (x+y)^2 FORMULA) NOW, X^2 AND Y^2 GET CANCELLED FROM BOTH =4-4x+16-(-8y)= 1-2x+9-(-6y) = 20-4x+8y=10-2x+6y = 10+2y=2x [taking x and y on different sides] DIVIDING THIS BY 2; = 5+y=x OR x=y+5 NOW WE HAVE X IN TERMS OF Y... USING AREA FORMULA OF COORDINATES OF A TRIANGLE, SUBSTITUTE THE VALUES,WRITE X AS 'Y+5' THEN IT BECOMES A ONE VARIABLE EQN EQUATE IT TO 9/2 AND U WILL GET 'Y' THEN FIND 'X' Sanya Mittal · 2 years, 8 months ago

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