In a $\Delta ABC$, the slope of the median through $A$ is $-2$, $B = (-1,3)$ and $C =(3,5)$. If its area is $5$ ....

Question

In a $\Delta ABC$, the slope of the median through $A$ is $-2$, $B = (-1,3)$ and $C =(3,5)$. If its area is $5$ then find the distance of the vertex $A$ from the origin.

I am not getting any idea how to proceed in such question, request you to please help. Thanks

Answer 1

Let $A(a,b)$ and the midpoint of BC is (1,4). Then, the slope of the median is $\frac{b-4}{a-1}=-2$, or,

$$2a+b=6\tag{1}$$

Use the area formula of the triangle given coordinates

$$5=\frac12|(x_A-x_C)(y_B-y_A)-(x_A-x_B)(y_C-y_A)|$$

to get

$$-a+2b=2\tag{2}$$

Solve (1) and (2) to obtain

$$a=2,\>\>\>b =2$$

Thus, the distance of the vertex A to the origin is $\sqrt{2^2+2^2}=2\sqrt{2}$

There is a second solution with the vertex A above BC, which can be obtained from the opposite sign of the absolute value of the area formula. The distance in the case is 6.

Answer 2

If we say $A = (a_1,a_2), B = (b_1,b_2), C = (c_1,c_2)$

By the shoelace algorithm the Area is

$\frac 12 |a_1b_2 - a_2b_1 + b_1c_2 - b_2c_1 + c_1a_2 - c_2a_1|$

With the numbers in this problem, that gives.

$\frac 12 |3a_1 + a_2 -5 - 9 + 3a_2 - 5a_1|\\ \frac 12 |-2a_1 + 4a_2 - 14| = 5\\ -2a_1 +4a_2 = 24 \text { or }-2a_1 +4a_2 = 4$

If you don't know the shoelace algorithm, you could say:

$B,C$ lie on the line $-x+2y = 7$

The distance between $B,C$ is $\sqrt {(3-(-1)^2 + (5-3)^2} = \sqrt {20}$

$\frac 12 h\times \sqrt 20 = 5\\ h = \sqrt 5$

The distance between lines $ax + by = c$ and $ax+by = d$ is $\frac {|c-d|}{\sqrt {a^2 +b^2}}$

The vertex must lie on either the line $-x + 2y = 12$ or $-x + 2y = 2$

the median cuts through the point $(\frac {b_1 + c_1}{2}, \frac {b_2+c_2}{2}) = (1,4)$

with slope -2, the line is

$y - 4 = -2(x-1)$

with $(a_1, a_2)$ falling on the line.

Giving us two systems of equations to solve.

$-x + 2y = 12\\ 2x + y = 6$

and

$-x +2y = 2\\ 2x + y = 6$

In fact, you can use that the slope of the median is perpendicular to the base to your advantage, stepping up 2 and over 1 above the line and below the line.

$(0,6), (2,2)$ are both candidates for $A.$

Answer 3

The median through $A$ goes through the midpoint $M$ of $BC$, too, which is located at $(1,4)$.
If the slope of $AM$ is $-2$ then $A(x,y)$ is located along the line with equation $y=-2x+6$.
By the shoelace formula

$$ 10 = |3x-5+3y+y-9-5x |=|4y-2x-14|=|-8x+24-2x-14|=|-10x+10|$$ hence $x\in\{0,2\}$ and $A$ is either located at $(0,6)$ or at $(2,2)$.
In the former case its distance from the origin is $6$, in the latter it is $2\sqrt{2}$.