Question

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).

Answer 1

1)

Gradient (m) of the line $=Tan45$

$=1$

Let $x=$ abscissa

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

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

$6+\frac{9}{2}=x+\frac{5}{2}$

$\implies\>x=8$

2)

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

The angle that the line makes with horizontal

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

$=57.26$

Angle that the line makes with any vertical line

$=90-57.26\\=32.74$

3)

Angle between the lines

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

Angle between $L_1$ and horizontal

$=Tan^{-1}(-1)\\=-45°$

Angle between $L_2$ and horizontal

$=-45+$ -$33.69$

$=-78.69$

Slope of $L_2=Tan(-78.69)$

$=-5$

Or between $L_2$ and horizontal $=-45+33.69$

$=-11.31$

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

$=-0.2$

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

$=26.565$

Angle between $L_2$ and horizontal

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

=3

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

$\sqrt{(-6)^2+3^2}=√45$

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

$=\sqrt{10^2+8^2)}=√164$

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

$=\sqrt{4^2+11^2}=√137$

Using cosine rule

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

$Cos\>\theta=\frac{256}{299.8}$

$\theta=31.36$