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Coordinate Geometry of the Straight Line - Questions

Question 1
17651

Given that \(A( - 3, - 12),\) \(\,B(0,3)\,\) and \(C(4,b)\) are collinear then the value of \(b\) is?

\(b = 23\)

\begin{align}
&m_{AB}=\frac{3-(-12)}{0-(-3)}=\frac{15}{3}=5\\
&m_{BC}=\frac{b-3}{4-0}=\frac{b-3}{4}\\
&\text{Collinear when } m_{AB}=m_{BC}\\
&\begin{aligned}
\rightarrow \;\frac{b-3}{4}&=5 \\
b&=23
\end{aligned}
\end{align}

Question 2
17652

The equation of the line that passes through the point \((1, - 1)\) and is perpendicular to the line \(y = 5x + 2\) is?

\(x + 5y + 4 = 0\)

\begin{align}
&m_{1}=5,\; m_{2}=-\frac{1}{5}\\
&\begin{aligned}
y-y_{1} &=m\left(x-x_{1}\right) \\
y-(-1) &=-\frac{1}{5}(x-1) \\
5(y+1) &=-(x-1) \\
5 y+5 &=-x+1
\end{aligned}\\
&\rightarrow x+5 y+4=0
\end{align}

Question 3
17654

The equation of the line that is parallel to the line \(3x - 2y + 1 = 0\) and passes through the point \({\rm{(}} - 1,3)\) is?

\(3x - 2y + 9 = 0\)

\begin{align}
&\text{The line will be of the form } 3x-2y+k=0\\
&(-1,3) \text{ lies on this line. }\\
&\therefore 3\times -1-2 \times 3+k=0 \rightarrow k=9\\
&\therefore 3 x-2 y+9=0
\end{align}

Question 4
17655

The line \(ax - 3y = 4\) is perpendicular to the line \(2x + 3y = 5.\)  The value of a is?

\(a = 4.5\)

\begin{align}
&\begin{aligned}
2x+3y &=5 \\
3y &=-2x+5 \\
y &=-\frac{2}{3} x+\frac{5}{3} \\
\therefore m_{1} &=-\frac{2}{3}
\end{aligned}\\
&\begin{aligned}
ax-3y &=4 \\
3y &=ax-4 \\
y&=\frac{a}{3} x-\frac{4}{3} \\
m_{2} &=\frac{a}{3}\\
\end{aligned}\\
&\begin{aligned}
\rightarrow \quad m_{1} m_{2}&=-1 \\
-\frac{2}{3} \times \frac{a}{3}&=-1 \rightarrow a=4.5
\end{aligned}
\end{align}

Question 5
17656

Find the value of \(p\) if the line passing through the points \((2p,\,p - 1)\) and \((p,\,3p + 1)\) makes an angle of \({\rm{4}}{{\rm{5}}^ \circ }\) with the positive direction of the horizontal axis. 

\(p = - \dfrac{2}{3}\)

\begin{align}
&m=\frac{3p+1-(p-1)}{p-2p}=\frac{2p+2}{-p}\\
&\begin{aligned}
m&=\tan \theta \\
&=\tan 45^{\circ}=1 \\
\end{aligned}\\
&\begin{aligned}
\therefore \frac{2p+2}{-p}&=1 \\
2p+2&=-p \\
3p&=-2 \\
p&=-\frac{2}{3}
\end{aligned}
\end{align}