Answer to Question #276198 in Geometry for Aliush

Question #276198
  1. A line with an angle of inclination of 45° passes through (-5/2 , -9/2). If the ordinate of the point is 6, what is its abscissa?
  2. Find the angle from the line through (13/2, 0) and (2, -7) to the vertical line 5 units from the y-axis.
  3. The angle from line 1 to line 2 is arctan (2/3) and the slope of line 1 is -1. Find the slope of line 2.
  4. The inclination of line 1 is arctan (1/2). If line 2 makes an angle of 45° with line 1, find the slope of line 2.
  5. The vertices of a triangle are (7, 4), (1, 7) and (-3, -4) Find the interior angle at the vertex (-3, -4).




1
Expert's answer
2021-12-07T11:03:15-0500

1)

Gradient (m) of the line =Tan45=Tan45

=1=1


Let x=x= abscissa

ΔYΔx=\frac{\Delta\>Y}{\Delta\>x}= gradient


6(92)x(52)=1\frac{6-(-\frac{9}{2})}{x-(\frac{-5}{2})}=1


6+92=x+526+\frac{9}{2}=x+\frac{5}{2}

     x=8\implies\>x=8



2)

Gradient of the line =702132=74.5=\frac{-7-0}{2-\frac{13}{2}}=\frac{7}{4.5}


The angle that the line makes with horizontal

=Tan1(74.5)=Tan^{-1}(\frac{7}{4.5})

=57.26=57.26


Angle that the line makes with any vertical line

=9057.26=32.74=90-57.26\\=32.74


3)

Angle between the lines

=Tan1(23)=33.69=Tan^{-1}(\frac{2}{3})=33.69


Angle between L1L_1 and horizontal

=Tan1(1)=45°=Tan^{-1}(-1)\\=-45°

Angle between L2L_2 and horizontal

=45+=-45+ -33.6933.69

=78.69=-78.69


Slope of L2=Tan(78.69)L_2=Tan(-78.69)

=5=-5

Or between L2L_2 and horizontal =45+33.69=-45+33.69

=11.31=-11.31


Slope of L2=L_2= Tan(11.31)Tan(-11.31)

=0.2=-0.2


4) Angle between L2andL1=Tan1(12)L_2\>and\,L_1=Tan^{-1}(\frac{1}{2})

=26.565=26.565

Angle between L2L_2 and horizontal

=45+26.565=71.565Slope=tan(71.565)=45+26.565=71.565\\ Slope= tan(71.565)

=3


5) side a=1774=\begin{vmatrix} 1-7 \\ 7-4 \end{vmatrix}


(6)2+32=45\sqrt{(-6)^2+3^2}=√45


Side b =7344=\begin{vmatrix} 7--3\\ 4--4 \end{vmatrix}


=102+82)=164=\sqrt{10^2+8^2)}=√164


Side c=1374\begin{vmatrix} 1--3\\ 7--4 \end{vmatrix}


=42+112=137=\sqrt{4^2+11^2}=√137

Using cosine rule


45=164+1372164×137cosθ45=164+137-2\sqrt{164×137}\> cos\theta


Cosθ=256299.8Cos\>\theta=\frac{256}{299.8}


θ=31.36\theta=31.36




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