Question

Without drawing the graphs , find the vertices of the triangles formed by the straight lines :

2x - y + 3 = 0 ; 4x - 3y + 5 = 0 and 5x - 3y + 10 = 0.

⭐CONTENT QUALITY REQUIRED⭐

Answer 1

By find the solution of two equations at once, We will be finding vertices.

First, 2x - y + 3 = 0 & 4x - 3y + 5 = 0 .

4x - 3y + 5 = 0
4x - 2y + 6 = 0
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-y -1 = 0
y = -1

Now, 4x +3 +5 = 0 , 4x +8 =0 , x = -2 .

One of the vertices of triangle = ( -2 , -1 )

Them, 4x - 3y + 5 = 0 & 5x - 3y + 10 = 0

4x - 3y + 5 = 0
5x - 3y + 10 = 0
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-x -5 0
x = -5

Now, -20 - 3y + 5 = 0 , -15 - 3y = 0 , -3y = 15 , y = -5

Another vertice of the triangle = ( -5 , -5)

At last, 2x - y + 3 = 0 , 5x - 3y + 10 = 0 .

6x -3y + 9 = 0
5x -3y + 10 = 0
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x - 1 = 0
x = 1 .

Now, 2 - y + 3 = 0 , 5 - y = 0 , y = 5 .

Third Vertices of the triangle = ( 1 , 5 )

Therefore ,The vertices of the triangle are ( 1 , 5 ) , ( -2 , -1 ) , ( -5 , -5 )

Answer 2

Answer :

We can see that these three lines intersect with each other ( i.e there exists a unique solution in any of the two lines ) by seeing that a1/a2 ≠ b1/b2 .

Given , three lines -

2x - y+3=0 ——— (i)

4x - 3y+5=0———(ii)

5x - 3y+10=0———(iii)

=>Solving equations ( i ) and (ii) simultaneously:-

( i ) ×2

4x - 2y + 6 = 0 (1)

Now, (1)-(ii),

We get :-

y+1=0

y=-1

Putting it in (i)

2x=-4

x= -2

One point of triangle => (-2,-1)

Now solving (ii) & (iii) simultaneously ,

(ii)-(iii), we get ->

x=-5

Putting this in (ii), we get:-

y= -5

2nd point of triangle => (-5,-5)

Now solving (iii) & (i) simultaneously =>

(i) × 3

we get ,

6x -3y + 9 = 0 ————(2)

Now, (2) - (iii), we get =>

x = 1

Putting it in (iii) , we get =>

y = 5

3rd vertex of triangle => (1,5)