# The sides of a triangle are formed by the lines with equations 2x-y-7=0, 3x+5y-4=0 and x+3y-4=0. a. Find vertices b.Find the exact length of all the altitudes of the triangle.

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To find the vertices of the triangle, we need to solve for the intersection of each pair of lines. This means that we need to solve the equations:

`2x-y-7=0` (1)

`3x+5y-4=0` (2)

`x+3y-4=0` (3)

And the vertices are at the solution of (1) and (2), then (1) and (3) and finally at (2) and (3).

**The first vertex is found by solving (1) and (2).** Multiply (1) by 5 then add to (2) to get

`13x-39=0`

which has solution `x=3` . Substitute back into (1) to get:

`6-y-7=0`

which has solution `y=-1`

The first vertex is `(3,-1)` .

**The second vertex is found by solving (1) and (3).** Multiply (1) by 3 then add to equation (3) to get:

`7x-25=0` which has solution `x=25/7` . Substitute into (1) to get

`50/7-y-7=0` which has solution `y=1/7` . The second vertex is `(25/7,1/7)` .

**The third vertex is found by solving (2) and (3).**

Multiply (3) by 3 then subtract from (2) to get:

`-4y+8=0` which has solution `y=2`

Substitute into (3) to get:

`x+6-4=0` which has solution `x=-2` . The third vertex is `(-2,2)` .

**The vertices are `(3,-1)` , `(25/7,1/7)` and `(-2,2)` .**

Let,

EQ1: `2x-y-7=0` EQ2: `3x+5y-4=0` EQ3: `x+3y-4=0`

(a) Vertices refer to the intersection point of two lines. This can be solved using elimination method.

> To determine the intersection point between EQ1 and EQ2, multiply EQ1 with 5.

`5*(2x-y-7)=0*5`

`10x-5y-35=0`

Then, add it with EQ.2

`10x-5y-35=0`

`(+)` `3x+5y-4=0`

`--------------`

`13x+0y - 39 = 0`

`x=3`

To solve for y, substitute x=13 to either EQ1 or EQ2.

`2(3)-y-7=0`

`y=-1`

*Hence, EQ1 and EQ2 intersects at point (3,-1).*

> To solve for the intersection point between EQ1 and EQ3, multiply EQ1 by 3.

`3(2x-y-7)=0*3`

`6x-3y-21=0`

Then, add it with EQ3.

`6x-3y-21=0`

`(+)` `x+3y-4=0`

`--------------`

`7x + 0y -25=0`

`x=25/7`

Then, substitute x=25/7 to either EQ1 or EQ2 and solve for y.

`2(25/7)-y-7=0`

`y=1/7`

*Thus, EQ1 and EQ3 intersects at point (25/7,1/7).*

> To determine the intersection point between EQ2 and EQ3, multiply EQ3 by 3.

`3(x+3y-4)=0*3`

`3x+9y-12=0`

Then, subtract it from EQ2.

`3x+5y-4=0`

`(-)` `3x+9y-12=0`

`--------------`

`0x-4y+8=0`

`y=2`

To solve for y, substitute y=2 to either EQ1 and EQ2.

`x+3(2)-4=0`

`x=-2`

*So, EQ2 and EQ3 intersects at point (-2,2) .*

**Therefore, the vertices of the triangle formed by the three equations are `(3,-1)` , `(25/7,1/7)` and `(-2,2)` .**

(b) Altitude is the perpendicular line from the vertex to its opposite side. It is also the minimum distance between the vertex and the line opposite to it.

So to solve for the length of the altitude, we may use the distance formula between a point and a line which is:

`d=|(Ax_1+By_1+C)/sqrt(A^2+B^2)|`

where A, B and C are the coefficients of the equation of the line and (x1,y1) is the point.

> The vertex formed by EQ1 and EQ2 is (3,-1), then its opposite side is EQ3.

So, substitute EQ3 and (3,-1) to the formula.

`d=|(3+3(-1)-4)/sqrt(1^2+3^2)| = |-4/sqrt10| = 4/sqrt10 = (2sqrt10)/5=1.27`

*Hence, the altitude between the vertex(3,-1) and its opposite side is 1.27 .*

> (25/7,1/7) is the vertex formed by EQ1 and EQ3. Then its opposite side is EQ2.

So we have,

`d=|(3(25/7)+5(1/7)-4) / sqrt(3^2+5^2)|= |(52/7)/sqrt34| =52/(7sqrt34) = (26sqrt34)/119 = 1.27`

*Hence, the altitude between the vertex (25/7,1/7) and its opposite side is 1.27.*

> The third vertex (-2,2) is formed by EQ2 and EQ3. Its opposite side is EQ1.

Susbtituting EQ1 and (-2,2) to the formula yields:

`d=|(2(-2)-2-7)/sqrt(2^2+(-1)^2) |= |-13/sqrt5|=13/sqrt5=(13sqrt5)/5 = 5.81`

*Thus, altitude between the vertex (-2,2) and its opposite side is 5.81.*

**Therefore, the length of the three altitudes of the triangle formed by given equations are 1.27, 1.27 and 5.81 .**

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