Question

Show that A (6, 4), B (5, -2) and C (7, -2) are the vertices of an isosceles triangle. Also find the length of the median through A.

Answer 1

Hint: First by using the distance formula we will find the length of the sides AB, BC and AC. Then if two sides of a given triangle are equal, then the vertices given are of isosceles triangle. Later to find the length of the median through A, so we find out the mid-point of BC using the mid-point formula then by using distance formula we will find the length of the median.

Formula used:
The distance between the two points using distance formula is,
\[d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\]Where \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] are the coordinates of the two given points.
Mid-point of any given line with two point is,
\[mid-po\operatorname{int}=\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)\], where \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] are the coordinates of the two given points.

Complete step by step answer:
As we know that,
The distance between the two points using distance formula is,
\[d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\] Where \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] are the coordinates of the two given points.
We have given the three vertices of the triangle i.e. A (6, 4), B (5, -2) and C (7, -2).

Now,
Using the distance formula,
Distance AB = \[\sqrt{{{\left( 5-6 \right)}^{2}}+{{\left( -2-4 \right)}^{2}}}=\sqrt{{{\left( -1 \right)}^{2}}+{{\left( -6 \right)}^{2}}}=\sqrt{1+36}=\sqrt{37}\]
Distance BC = \[\sqrt{{{\left( 7-5 \right)}^{2}}+{{\left( -2-\left( -2 \right) \right)}^{2}}}=\sqrt{{{\left( -2 \right)}^{2}}+{{\left( 0 \right)}^{2}}}=\sqrt{4}=2\]
Distance AC = \[\sqrt{{{\left( 7-6 \right)}^{2}}+{{\left( -2-4 \right)}^{2}}}=\sqrt{{{1}^{2}}+{{\left( -6 \right)}^{2}}}=\sqrt{1+36}=\sqrt{37}\]
Therefore,
AB = AC = \[\sqrt{37}\].
Thus, triangle ABC is an isosceles triangle as the two sides of the triangles are equal.
Therefore, A (6, 4), B (5, -2) and C (7, -2) are the vertices of an isosceles triangle.
Now,
Median use the line segment joining the vertex to the midpoint of the opposite side.
Therefore, median from A to the mid-point of BC.
Using the mid-point formula i.e. \[mid-po\operatorname{int}=\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)\], where \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\]are the coordinates of the two given points.
Mid-point of BC = \[\left( \dfrac{5+7}{2},\dfrac{-2+\left( -2 \right)}{2} \right)=\left( \dfrac{12}{2},\dfrac{-4}{2} \right)=\left( 6,-2 \right)\]
Therefore, the mid-point of BC is D having the co-ordinate (6, -2).

Using the distance formula,
Distance AD = \[\sqrt{{{\left( 6-6 \right)}^{2}}+{{\left( -2-4 \right)}^{2}}}=\sqrt{{{\left( -6 \right)}^{2}}}=\sqrt{36}=6\]
Therefore the length of the median through A is 6.

Note: In order to solve the given question, students need to remember the concepts of triangle and the types of triangle. They should be well aware of the formula of finding the distance between two points. They should also know about the concept of median and how to find the mid-point of any given line with two points. Solve the questions very carefully and explicitly to avoid making any error.