Check whether (5, - 2), (6, 4) and (7, - 2) are the vertices of an isosceles triangle

We know that in an isosceles triangle, length of two sides is equal.

We know that formula to find the distance (d) between two points $\left( {{x}_{1}},{{y}_{1}} \right)$and $\left( {{x}_{2}},{{y}_{2}} \right)$ is:

$d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$

Let, A = (5, -2), B = (6, 4) and C = (7, -2).

The distance between A and B is:

$\begin{align}

  & \text{AB}=\sqrt{{{\left( 6-5 \right)}^{2}}+{{\left( 4+2 \right)}^{2}}} \\

 & =\sqrt{1+36} \\

 & =\sqrt{37} 

\end{align}$

The distance between B and C is:

$\begin{align}

  & \text{BC}=\sqrt{{{\left( 7-6 \right)}^{2}}+{{\left( -2-4 \right)}^{2}}} \\

 & =\sqrt{1+36} \\

 & =\sqrt{37} 

\end{align}$

The distance between C and A is:

$\begin{align}

  & \text{CA}=\sqrt{{{\left( 7-5 \right)}^{2}}+{{\left( -2+2 \right)}^{2}}} \\

 & =\sqrt{4+0} \\

 & =2 

\end{align}$ We get that the length of side AB and BC are equal. So, the given points are the vertices of an isosceles triangle.

Question 3 Coordinate Geometry - Exercise 7.1

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Answer:

Since two sides of any isosceles triangle are equal. To check whether given points are vertices of an isosceles triangle, we will find the distance between all the points.

Let the points (5, – 2), (6, 4), and (7, – 2) are representing the vertices A, B, and C respectively.

A B =\sqrt{(6-5)^{2}+(4+2)^{2}}=\sqrt{(-1)^{2}+(6)^{2}}=\sqrt{37} \\ B C =\sqrt{(7-6)^{2}+(-2-4)^{2}}=\sqrt{(-1)^{2}+(6)^{2}}=\sqrt{37} \\ C A =\sqrt{(7-5)^{2}+(-2+2)^{2}}=\sqrt{(-2)^{2}+(0)^{2}}=2 \\ \text { Here } A B=B C =\sqrt{37}

This implies that the given points are vertices of an isosceles triangle.

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Find the coordinates of the points which divide the line segment joining A(– 2, 2) and B(2, 8) into four equal parts.

Let P, Q and R be the three points which divide the line-segment joining the points A(-2, 2) and B(2, 8) in four equal parts.

Case I. For point P, we have

Hence, m1 = 1, m2 = 3
x1 = -2, y2 = 2
x2 = 2, y2 = 8Then, coordinates of P are given by

Case II. For point Q, we have

m1 = 2, m2 = 2
x1 = -2, y1 = 2
and    x2 = 2, y2 = 8Then, coordinates of Q are given by

Case III. For point R, we have

Check whether 5, 2, 6, 4 and 7, 2 are the vertices of an isosceles triangle.

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Solution:

An isosceles triangle is a triangle that has two sides of equal length.

To check whether the given points are vertices of an isosceles triangle, we need to check the distance between any of the 2 points should be the same for two pairs of given points.

Let the points (5, - 2), (6, 4), and (7, - 2) represent the vertices A, B, and C of the given triangle

We know that the distance between the two points is given by the Distance Formula,

Distance Formula =  √[(x₂ -  x₁)2 + (y₂ - y₁)2]

To find AB, that is distance between points A (5, - 2) and B (6, 4), let x₁ = 5, y₁ = -2, x₂ = 6, y₂ = 4

AB = √[( 6 - 5 )2 + (4 - (-2))2]

= √[(1)2 + (6)2]

= √1 + 36

= √37

To find BC, distance between Points B (6, 4) and C (7, - 2), let x₁ = 6, y₁ = 4, x₂ = 7, y₂ = - 2

BC = √ [( 7 - 6 )2 + (-2 - 4)2]

= √[(1)2 + (- 6)2]

= √1 + 36

= √37

To find AC, that is distance between Points A (5, - 2) and C (7, - 2), let x₁ = 5, y₁ = - 2, x₂ = 7, y₂ = - 2

AC = √ [( 7 - 5 )2 + (-2 - (- 2))2]

= √[(2)2 + (0)2]

= 2

From the above values of AB, BC and AC we can conclude that AB = BC. As the two sides are equal in length, therefore, ABC is an isosceles triangle.

☛ Check: NCERT Solutions for Class 10 Maths Chapter 7

Video Solution:

NCERT Class 10 Maths Solutions Chapter 7 Exercise 7.1 Question 4

Summary:

The points (5, - 2), (6, 4), and (7, - 2) are the vertices of an isosceles triangle.

☛ Related Questions:

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