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 )
[tex]\text{slope = }\frac{y_2-y_1}{x_2-x_1}=\frac{2-(-7)}{8-\text{ (-4)}}=\frac{9}{12}=\frac{3}{4}[/tex]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
[tex]\text{slope}=-\frac{4}{3}[/tex]The point-slope form of a line is :
[tex]y-y_1=m(x-x_1)[/tex]where m is the slope = -4/3
at point C ( -6, 8)
[tex]\begin{gathered} y-8\text{ = -}\frac{4}{3}(x-(-6)) \\ y-8=-\frac{4}{3}(x+6) \end{gathered}[/tex]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:
[tex]\begin{gathered} y-(-7)\text{ = }\frac{3}{4}(x-(-4)) \\ y+7=\frac{3}{4}(x+4) \\ 4\cdot\lbrack y+7\text{ = }\frac{3}{4}(x+4)\rbrack\cdot4 \\ y+28=3(x+4) \\ y+28=3x+12 \\ 3x-y-16=0 \end{gathered}[/tex]Equation of the line that passes through C:
[tex]\begin{gathered} y-8=-\frac{4}{3}(x+6) \\ 3\cdot\lbrack y-8=-\frac{4}{3}(x+6)\rbrack\cdot3 \\ y-24=-4(x+6) \\ y-24=-4x+24 \\ 4x+y-48=0 \end{gathered}[/tex]The point of intersection of line
[tex]\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}[/tex][tex]\begin{gathered} AB\text{ : }3x-y-16=0\text{ } \\ a_1=3 \\ b_1=-1 \\ c_1=-16 \end{gathered}[/tex][tex]\begin{gathered} C\colon\text{ }4x+y-48=0 \\ a_2=4 \\ b_2=1 \\ c_2=-48 \end{gathered}[/tex]The point of intersection, D:
[tex]\begin{gathered} D(x,y)=\lbrack\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1},\frac{a_2c_1-a_1c_2}{a_1b_2-a_2b_1}\rbrack \\ D(x,y)=\lbrack\frac{(-1)(-48)-(1)(-16)}{(3)(1)-(4)(-1)},\frac{(4)(-16)-(3)(-48)}{(3)(1)-(4)(-1)}\rbrack \\ D(x\mathrm{}y)=\lbrack\frac{48+16}{3+4},\frac{-64_{}+144}{3+4}\rbrack=\lbrack\frac{64}{7},\frac{80}{7}\rbrack \\ D(xy)=D(\frac{64}{7},\frac{80}{7}) \end{gathered}[/tex]d) the kind of triangle CBD and CAD
Both triangles are right triangle , since sline CD is perpendicular to line AB
