Question

In the following diagram triangle ABC is drawn, with A(-7,-4), B(8,2) and C(-6,8).(a) determine the slope of line AB in simplest form.(b) draw the line that passes through C and is perpendicular to line AB. Write it's equation in point-slope form (c) If point D lies at the intersection of line AB with the line you drew in (b), mark D and state its coordinates.(d) what kinds of triangles are triangle CBD and CAD? Why)

Answer 1

Given are the points of the vertices of the triangle.

A ( -4 , -7 )

B ( 8, 2 )

C ( -6, 8 )

Required:

a) slope of line AB

b) equation of the line that passes through C and is perpedicular to AB

c) coordinates of D which is the intersection of AB and the line that passes through C

d) the kind of triangle CBD and CAD

Solution :

a) slope of line AB

A ( -4 , -7 )

B ( 8, 2 )

`\text{slope = }(y_2-y_1)/(x_2-x_1)=\frac{2-(-7)}{8-\text{ (-4)}}=(9)/(12)=(3)/(4)`

b) equation of the line that passes through C and is perpedicular to AB.

The slopes of two perpendicular lines are negative reciprocals of each other. Therefore, the slope of the line that passes through C ( -6, 8 ) is

`\text{slope}=-(4)/(3)`

The point-slope form of a line is :

`y-y_1=m(x-x_1)`

where m is the slope = -4/3

at point C ( -6, 8)

`\begin{gathered} y-8\text{ = -}(4)/(3)(x-(-6)) \\ y-8=-(4)/(3)(x+6) \end{gathered}`

c) coordinates of D which is the intersection of AB and the line that passes through C

Equation of line AB : Let's take point A ( -4 , -7 ) and slope is 3/4 . Using the point intercept form:

`\begin{gathered} y-(-7)\text{ = }(3)/(4)(x-(-4)) \\ y+7=(3)/(4)(x+4) \\ 4\cdot\lbrack y+7\text{ = }(3)/(4)(x+4)\rbrack\cdot4 \\ y+28=3(x+4) \\ y+28=3x+12 \\ 3x-y-16=0 \end{gathered}`

Equation of the line that passes through C:

`\begin{gathered} y-8=-(4)/(3)(x+6) \\ 3\cdot\lbrack y-8=-(4)/(3)(x+6)\rbrack\cdot3 \\ y-24=-4(x+6) \\ y-24=-4x+24 \\ 4x+y-48=0 \end{gathered}`

The point of intersection of line

`\begin{gathered} a_1x+b_1y+c_1=0\text{ } \\ \text{and} \\ a_2x+b_2y+c_2=0\text{ } \\ is \\ (x,y)=\lbrack\frac{b_1c_2-b_2c_1}{a_1_{}b_2-a_2b_1},\frac{a_2c_1-a_1_{}c_2}{a_1b_2-a_2b_1}\rbrack \end{gathered}`
`\begin{gathered} AB\text{ : }3x-y-16=0\text{ } \\ a_1=3 \\ b_1=-1 \\ c_1=-16 \end{gathered}`
`\begin{gathered} C\colon\text{ }4x+y-48=0 \\ a_2=4 \\ b_2=1 \\ c_2=-48 \end{gathered}`

The point of intersection, D:

`\begin{gathered} D(x,y)=\lbrack(b_1c_2-b_2c_1)/(a_1b_2-a_2b_1),(a_2c_1-a_1c_2)/(a_1b_2-a_2b_1)\rbrack \\ D(x,y)=\lbrack((-1)(-48)-(1)(-16))/((3)(1)-(4)(-1)),((4)(-16)-(3)(-48))/((3)(1)-(4)(-1))\rbrack \\ D(x\mathrm{}y)=\lbrack(48+16)/(3+4),\frac{-64_{}+144}{3+4}\rbrack=\lbrack(64)/(7),(80)/(7)\rbrack \\ D(xy)=D((64)/(7),(80)/(7)) \end{gathered}`

d) the kind of triangle CBD and CAD

Both triangles are right triangle , since sline CD is perpendicular to line AB