Need help to verify solution

[westin]

A triangle has vertices A(−1, 2), B(5, 8) and C(−1, −7). Point D(x, y) is on side BC, and the area of triangle ACD is half the area of triangle ABC. What is the value of x + y?

Hi, the official answer for the above is 2. However, I got 2.5. Am I missing something or the official answer is not correct. Can someone help to verify?

Area of ABC = 27 [(Base x height)/2 = (9 x 6)/ 2 = 27)
Area of ACD = 13.5 (half of area of ABC)
SInce base of ABD is 9 and height of ABD is (x+1), then 9(x+1)/2 = 13.5 => x = 2
Find the linear equation of BC using the two coordinates, we got equation: y= 5/2x - 9/2. using x =2, y = 0.5

As a result, my answer of x + y = 2.5. however, the official answer is 2.

Can someone verify my answer pls. thanks!!!!

btw, is there a better free graphing calculator that can plot shape just using coordinates instead of equation. I ma using desmos in a cumbersome way...

 

[Dr.Peterson]

A triangle has vertices A(−1, 2), B(5, 8) and C(−1, −7). Point D(x, y) is on side BC, and the area of triangle ACD is half the area of triangle ABC. What is the value of x + y?

Hi, the official answer for the above is 2. However, I got 2.5. Am I missing something or the official answer is not correct. Can someone help to verify?

Area of ABC = 27 [(Base x height)/2 = (9 x 6)/ 2 = 27)
Area of ACD = 13.5 (half of area of ABC)
SInce base of ABD is 9 and height of ABD is (x+1), then 9(x+1)/2 = 13.5 => x = 2
Find the linear equation of BC using the two coordinates, we got equation: y= 5/2x - 9/2. using x =2, y = 0.5

As a result, my answer of x + y = 2.5. however, the official answer is 2.

Can someone verify my answer pls. thanks!!!!

I agree with you.

A quicker solution is to see that D must be the midpoint of BC, which is ((5-1)/2, (8-7)/2) = (2, 1/2).

btw, is there a better free graphing calculator that can plot shape just using coordinates instead of equation. I ma using desmos in a cumbersome way...

Have you tried GeoGebra? That does many things Desmos can't, though Desmos beats it in some other tasks. It's much more than a "graphing calculator".

 

[westin]

Thank you Dr. Peterson again! Great to know that my answer is correct. My first instinct also thinks that the midpoint of BC is D too, however, I do not know how to prove it theoretically that is true. Can you help me more to realize that why half of the area meant that D is the midpoint. Is there a geometry theorem that states that?

 

[westin]

wow, GeoGebra is much better than Desmos. thank you very much for the recommendation!

 

[Otis]

base of ABD is 9 and height of ABD is (x+1)

You'd meant to write ACD, Westin, and I like your reasoning.

I used an offshoot of Heron's Formula to confirm D(2,1/2) -- it uses vertex coordinates.

(I'd derived that offshoot formula so that I could program my calculator to do grunt work, but I still use it in paper work.)

(a,b)
(c,d)
(e,f)

A = | (1/2)(a-c)(d-f) - (1/2)(c-e)(b-d) |

?

 

[Dr.Peterson]

Thank you Dr. Peterson again! Great to know that my answer is correct. My first instinct also thinks that the midpoint of BC is D too, however, I do not know how to prove it theoretically that is true. Can you help me more to realize that why half of the area meant that D is the midpoint. Is there a geometry theorem that states that?

This is a result of the fact that the areas of triangles with the same altitude are proportional to their bases. Take BC and CD as the bases, and A as the vertex to which the altitude is measured.

If they had said D was on the line BC rather than the segment (side) BC, there would be a second solution, the same distance on the other side of C.

 

[westin]

I see it now. Since ABD and ACD have same area and same height, then the base of that two triangle has to be equal. Meaning D is the midpoint. Thank you again for the help.