Question

Plot the points (6,5).(6,-3) and (-2,-3) join them to find the figure​

Answer 1

SOLUTION

TO DETERMINE

Plot the points (6,5).(6,-3) and (-2,-3) join them to find the figure

EVALUATION

Here the given points are (6,5).(6,-3) and (-2,-3)

We denote them as below

A(6,5) , B (6,-3) and C (-2,-3)

Figure : Figure is referred to the attachment

We plot them in the figure.

We denote the line AB by Red colour

We denote the line BC by Green colour

We denote the line AC by Blue colour

After plotting the points and adding them we get a triangle

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Answer 2

Given :

A = (6 , 5)

B = (6 , -3)

C = (-2 , -3)

To find :

Plot the points (6,5) , (6,-3) and (-2,-3) join them to find the figure​.

Solution :

$Distance = \sqrt{(x_2 - x_1)^{2} +(y_2-y_1)^{2} }$

Distance between point A and B

$d_a_b = \sqrt{(6 - 6)^{2} +(-3-5)^{2} }$

$d_a_b = \sqrt{(-8)^{2} }$

$d_a_b = 8$

Distance between point A and B

$d_b_c = \sqrt{(-2-6)^{2} +(-3-(-3))^{2} }$

$d_b_c = \sqrt{(-8)^{2} + (-3+3 )^{2} }$

$d_b_c = \sqrt{(-8)^{2} }$

Distance between point C and A

$d_c_a = \sqrt{(6-(-2))^{2} + (5-(-3))^{2} }$

$d_c_a = \sqrt{(6+2)^{2} +(5 + 3)^{2} }$

$d_c_a =\sqrt{8^{2} +8^{2} }$

$d_c_a = 8\sqrt{2}$

After joining the points we can see that it is an isosceles triangle because

two of the distances are equal.

$tan\theta = \frac{perpendicular}{base}$

tanθ = 8 / 8

tanθ = 1

θ = 45°

∠BCA = 45°

∵ It is an isosceles triangle.

∴ ∠BAC = 45°

Sum of all the angles of the triangle is equal to 180°

∴ ∠ABC + ∠BCA + ∠CAB = 180°

∠ABC + 45° + 45° = 180°

∠ABC + 90° = 180°

∠ABC = 180° - 90°

∠ABC = 90°

∵ One of the angle is 90°

∴ The triangle is right angled triangle.

So, we can say that after joining all the three points we get an isosceles right angled triangle.