Question

$\bf Question..!!$

The area of a triangle is 5 square units, two of its vertices are (2, 1) and (3, -2). The third vertex lies on y = x+ 3. What will be the third vertex ?

- Don't spam.!
- No irrelevant answers..! ​

Answer 1

Answer:

We will assume the coordinate of the third vertex to be (x,y)

.

It is also given that the third vertex lies on y=x+3

………….. (1)

We will now write the formula of the area of the triangle in terms of the vertex of the triangle.

So, area of the triangle =12[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]

Now, by equating this area of the triangle is equal to the 5 square units as given in the

question. Therefore, we get

⇒12[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]=5

On cross multiplication, we get

⇒[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]=10

Now substituting the values of the coordinates in the above equation, we get

⇒[2(−2−y)+3(y−1)+x(1+2)]=10

⇒−4−2y+3y−3+3x=±10

By simplifying the above equation, we get

⇒y+3x=17

…………… (2)

⇒y+3x=−3

…………… (3)

So by solving the equation (1) and equation (2), we get

⇒x+3+3x=17

Adding the like terms, we get

⇒4x=14

Dividing both side by 4, we get

⇒x=72

And by solving the equation (1) and equation (3), we get

⇒x+3+3x=−3

Adding the like terms, we get

⇒4x=−6

Dividing both side by 4, we get

⇒x=−32

Now by putting the value of x

in equation (1) we will get the value of y

. Therefore, we get

y=72+3=132

for corresponding to the value of x=72

y=−32+3=32

for corresponding to the value of x=−32

So, (72,132)

or(−32,32)

is the coordinates of the third vertex.

Answer 2

$\large\underline{\sf{Solution-}}$

Let assume that the required triangle be ABC having area 5 square units and Coordinates of A be (2, 1) and Coordinates of B be (3, -2).

Since it is given that third Coordinates lies om the line y = x + 3.

So, Let assume that Coordinates of C be (h, k)

As, it is given that (h, k) satisfied y = x + 3

$\rm\implies \:k = h + 3$

So,

$\rm\implies \:Coordinates \: of \: C = (h, \: h + 3)$

So, we have now

Coordinates of A (2, 1)

Coordinates of B (3, - 2)

Coordinates of C (h, h + 3)

Area of triangle ABC = 5 square units

We know,

If A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of a triangle, then the area of triangle is given by

$\begin{gathered}\boxed{\tt{ Area =\dfrac{1}{2}\bigg|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\bigg|}} \\ \end{gathered}$

So, on substituting the values, we get

$\rm \: 5 =\dfrac{1}{2}\bigg|2( - 2-h - 3)+3(h + 3-1) + h(1 + 2)\bigg|$

$\rm \: 10 =\bigg|2( - 5-h)+3(h + 2) + 3h\bigg|$

$\rm \: 10 =\bigg|- 10-2h+3h + 6 + 3h\bigg|$

$\rm \: 10 =\bigg|4h - 4\bigg|$

$\rm\implies \:4h - 4 = 10 \: \: or \: 4h - 4 = - 10$

$\rm\implies \:4h = 14 \: \: or \: 4h = - 6$

$\rm\implies \:h = \dfrac{7}{2} \: \: or \: \: h = - \dfrac{3}{2}$

So,

$\rm \: Coordinates \: of \: C = (h, \: h + 3)$

On substituting the value of h, we get

$\rm \: Coordinates \: of \: C = \bigg( \dfrac{7}{2} , \: \dfrac{7}{2} + 3\bigg) \: \: or \: \: \bigg( - \dfrac{3}{2} , \: - \dfrac{3}{2} + 3\bigg)$

$\boxed{\tt{ \rm \: Coordinates \: of \: C = \bigg( \dfrac{7}{2} , \: \dfrac{13}{2} \bigg) \: \: or \: \: \bigg( - \dfrac{3}{2} , \: \dfrac{3}{2}\bigg)}}$

is the required coordinate.

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

LEARN MORE

1. Distance Formula

If A(x₁, y₁) and B(x₂, y₂) be two points in the cartesian plane, then distance between A and B is given by

$\begin{gathered}\boxed{\tt{ AB \: = \sqrt{ {(x_{1} - x_{2}) }^{2} + {(y_{2} - y_{1})}^{2} }}} \\ \end{gathered}$

2. Section formula

If A(x₁, y₁) and B(x₂, y₂) be two points in the cartesian plane and C(x, y) be the point which divides AB internally in the ratio m₁ : m₂, then the coordinates of C is given by

$\begin{gathered} \boxed{\tt{ (x, y) = \bigg(\dfrac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}, \dfrac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}\bigg)}} \\ \end{gathered}$

3. Mid-point formula

If A(x₁, y₁) and B(x₂, y₂) be two points in the coordinate plane and C(x, y) be the mid-point of AB, then the coordinates of C is given by

$\begin{gathered}\boxed{\tt{ (x,y) = \bigg(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\bigg)}} \\ \end{gathered}$

4. Centroid of a triangle

Centroid of a triangle is defined as the point at which the medians of the triangle meet and is represented by the symbol G.

If A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of a triangle and G(x, y) be the centroid of the triangle, then the coordinates of G is given by

$\begin{gathered}\boxed{\tt{ (x, y) = \bigg(\dfrac{x_{1}+x_{2}+x_{3}}{3}, \dfrac{y_{1}+y_{2}+y_{3}}{3}\bigg)}} \\ \end{gathered}$

5. Condition for 3 points to be Collinear

If A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the coordinates in cartesian plane, then points A, B and C are collinear, then

$\begin{gathered}\boxed{\tt{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) = 0}} \\ \end{gathered}$