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# Find the third vertex of a triangle if its two vertices are $\left( { - 1,4} \right)$ and $\left( {5,2} \right)$ and midpoint of one side is $\left( {0,3} \right)$.

Last updated date: 18th Jun 2024
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Answer
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Hint: Here we will first assume the third vertex. We will then use the midpoint formula and substitute the values of vertices in the formula. We will simplify it to find the third vertex. We will take the second condition and solve it using the midpoint formula to find other possible solutions.

Formula used:
Mid-point $\left( {a,b} \right)$ formula of a line segment for the two vertices $\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right)$, we get
$\dfrac{{{x_1} + {x_2}}}{2} = a,\dfrac{{{y_1} + {y_2}}}{2} = b$.

Complete step by step solution:
Given two vertices of a triangle are $A\left( { - 1,4} \right)$ and $B\left( {5,2} \right)$.
It is given that point $\left( {0,3} \right)$ is the midpoint of one side.
Let the third vertex of the triangle be $C\left( {x,y} \right)$.
So, by using the given information, we can draw the triangle as:

Now they form the two conditions when the midpoint $\left( {0,3} \right)$ is the midpoint of side AC or the mid-point is of side BC.
Firstly taking the midpoint $\left( {0,3} \right)$ be the midpoint of the side AC.
Now by using the concept of the midpoint formula for the two vertices $\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right)$, we get
$\dfrac{{{x_1} + {x_2}}}{2} = 0$ and $\dfrac{{{y_1} + {y_2}}}{2} = 3$
Substituting ${x_1} = x$, ${x_2} = - 1$, ${y_1} = y$ and ${y_2} = 4$ in the above equations, we get
$\Rightarrow \dfrac{{x + \left( { - 1} \right)}}{2} = 0$ and $\dfrac{{y + 4}}{2} = 3$
Simplifying the expression, we get
$\Rightarrow x - 1 = 0$ and $y + 4 = 6$
$\Rightarrow x = 1$ and $y = 2$
Hence the third vertex is $\left( {1,2} \right)$.
Now we will take the second condition i.e. mid-point $\left( {0,3} \right)$ is the midpoint of the side BC.
Now by using the basic concept of the midpoint formula, we get
$\Rightarrow \dfrac{{x + 5}}{2} = 0$ and $\dfrac{{y + 2}}{2} = 3$
On cross multiplication, we get
$\Rightarrow x + 5 = 0$ and $y + 2 = 6$
Adding and subtracting the like terms, we get
$\Rightarrow x = - 5$ and $y = 4$
Hence the third vertex is $\left( { - 5,4} \right)$.

Hence the two possible third vertices are $\left( {1,2} \right)$ and $\left( { - 5,4} \right)$.

Note:
Mid-point is the middle point of a line segment i.e. it divides the line segment into the two equal halves. Trisect means that the line segment is divided into three parts and all the three parts are equal i.e. equally divided. Coordinates system is represented in the Cartesian plane and the coordinates are written in such a way that the X intercept is written firstly and then the Y intercept is written after X coordinate in the form of $\left( {x,y} \right)$.