Question

Find the equations of the sides of a triangle whose vertices are (3, –5),

(1, 2) and (7, –1).​

Answer 1

Question : -

Find the equations of the sides of a ∆ whose vertices are (3,-5),(1,2) & (7,-1)?

ANSWER

Given : -

Vertices of ∆ are (3,-5),(1,2) & (7,-1)

Required to find : -

• Equations of the sides of ∆ ?

Formula used : -

Slope of the line (m)

Slope (m) = (change in y co-ordinates)/(change in x co-ordinates)

Equation of the line is y = mx + b

Here,

m = slope of the line

x = Any one of x co-ordinate of the point

b = y intercept it is a point where the line intersects the y-axis

Solution : -

Vertices of ∆ are (3,-5),(1,2) & (7,-1)

So,

Let the ∆ be ABC

In ∆ABC,

• A = (3,-5)
• A = (3,-5)B = (1,2)
• A = (3,-5)B = (1,2)C = (7,-1)

The sides of ∆ ABC are;

AB,BC & AC

Now,

Consider the side AB

Vertices of side AB are A(3,-5) & B(1,-2)

Here,

slope of the line = (2-[-5])/(1-3)

slope of the line = (2+5)/(-2)

slope of the line = (7)/(-2)

• slope of the line AB = (-7)/(2)

Equation of line is y = mx+b

This implies;

y = (-7)/(2)x+b ..........(1)

choose any y & x co-ordinate of line AB in order to find the value of b

-5 = (-7)/(2)(1) + b

-5 = (-7)/(2) + b

-5 - (-7)/(2) = b

-5 + (7)/(2) = b

(-10+7)/(2) = b

(-3)/(2) = b

b = (-3)/(2)

substituting the value of b in eq-1

y = (-7x)/(2)+(-3)/(2)

y - (-7x)/(2) = (-3)/(2)

y + (7x)/(2) = (-3)/(2)

(7x)/(2) + y = (-3)/(2)

Hence,

• Equation of side AB = (7x)/(2)+y = (-3)/(2)

Now,

Consider side BC

The vertices of side BC are B(1,2) & C(7,-1)

Here,

Slope of the line = (-1-2)/(7-1)

Slope of the line = (-3)/(6)

slope of the line = (-1)/(2)

• slope of line BC = (-1)/(2)

Equation of line BC =

y = mx + b

y = (-1)/(2)x + b ....... (2)

This implies;

2 = (-1)/(2)(2) + b

2 = (-1) + b

2 - (-1) = b

2 + 1 = b

3 = b

b = 3

Substituting the value of b in Equation 2

y = (-x)(2) + 3

y - (-x)/(2) = 3

y + (x)/(2) = 3

(x)/(2) + y = 3

Hence,

• Equation of side BC = (x)/(2)+y = 3

At last,

Consider side AC

The vertices of side AC are A(3,-5) & C(7,-1)

Here,

Slope of the line = (-1-[-5])/(7-3)

Slope of the line = (-1+5)/(4)

slope of the line = (4)/(4)

• slope of line BC = 1

Equation of line BC =

y = mx + b

y = (1)x + b ....... (3)

This implies;

-5 = (1)(3) + b

-5 = 3 + b

-5 - 3 = b

-8 = b

b = -8

Substituting the value of b in Equation 3

y = x + (-8)

y - x = -8

-x+y = -8

x-y = 8

Hence,

• Equation of side AC = x-y = 8

Therefore,

Equation of sides ∆ABC are;

• AB is (7x)/(2)+y = (-3)/(2)
• BC is (x)/(2)+y = 3
• AC is x-y = 8

Answer 2

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