Straight Lines

30.

Assuming L along y-axis and C along x-axis, we have two points (124.942, 20) and (125.134, 110) in xy-plane. By two-point form, the point L and C satisfies the equation.

y-124.942= $\frac{(125.134-124.942)}{(110-20)}$  (x-20)

$\Rightarrow$ y-124.942= $\frac{0.192}{90}$  (x - 20)

$\Rightarrow$ y-124.942= $\frac{0.032}{15}$  (x - 20)

$\Rightarrow$ 15y – 1874.13=0.0

29.

The slope of line passing through (0, 0) and (–2, 9) is

$m_1=\frac{y_2-y_1}{x_2-x_1}=\frac{9-0}{-2-0}=\frac{9}{-2}$

The line perpendicular to the line having slope m1 will have the slope

$m_2=\frac{-1}{m_1}=\frac{-1}{-9/2}=\frac{2}{9}$

So, the equation of line with slope m2 and passing $\left(x_0,y_0\right)=\left(-2,9\right)$ is

$m_2=\frac{y-y_0}{x-x_0}$

$\Rightarrow \frac{2}{9}=\frac{y-9}{x-\left(-2\right)}=\frac{y-9}{x+2}$

$\Rightarrow 2\left(x+2\right)=9\left(y-9\right)$

$\Rightarrow 2x+4=9y-81$

$\Rightarrow 2x-9y+4+81=0$

$\Rightarrow 2x-9y+85=0$

Let a and b be the intercepts on x and y-axis

Then a + b = 9

$\Rightarrow b=9-a\left(1\right)$

Using intercept form equation of line with a and b intercepts are

$\frac{x}{a}+\frac{y}{b}=1$

$\Rightarrow \frac{x}{a}+\frac{y}{9-a}=1\left(2\right)$

As point (2, 2) passes the line having equation of form equation (2) we can write as

$\frac{2}{a}+\frac{2}{9-a}=1$

$\Rightarrow \frac{2\left(9-a\right)+2a}{a\left(9-a\right)}=1$

$\Rightarrow 18-2a+2a=\left(9-a\right)a$

$\Rightarrow 18=9a-a^2$

$\Rightarrow a^2-9a+18=0$

$\Rightarrow a^2-3a-6a+18=0$

$\Rightarrow a\left(a-3\right)-6\left(a-3\right)=0$

$\Rightarrow \left(a-3\right)\left(a-6\right)=0$

So, a = 3, 6

Case I

When a = 3, b = 9 – 3 = 6. Then the equation of line is

$\frac{x}{3}+\frac{y}{c}=1$

$\Rightarrow \frac{2x+y}{6}=1$

$\Rightarrow 2x+y=6$

$\Rightarrow 2x+y-6=0$

Case II

When a = 6,

26.

Let the line cuts x and y axis at intercept c. Then a = b = c

Then by intercept form the equation of line is

$\frac{x}{a}+\frac{y}{b}=1$

$\Rightarrow \frac{x}{c}+\frac{y}{c}=1$

$\Rightarrow x+y=c\left(1\right)$

Since equation (1) passes through point (2, 3).

Hence 2 + 3 = c

$\Rightarrow c=5$

So, substituting value of c in equation (1) we get

$\Rightarrow x+y=5$

$\Rightarrow x+y-5=0$

25.

Let P(1, 0) and Q(2, 3) be the given points. Then, slope of line joining PQ,

$m_1=\frac{3-0}{2-1}=\frac{3}{1}=3$

If A divides PQ with ratio 1:n, co-ordinate of A is

$\left(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n}\right)$

$=\left(\frac{1\times 2+n\times 1}{1+n},\frac{1\times 3+n\times 0}{1+n}\right)$

$=\left(\frac{2+n}{1+n},\frac{3}{1+n}\right)$

So, slope of line perpendicular to line joining points P and Q

$m_2=\frac{-1}{m_1}=-\frac{1}{3}$

Hence, equation of line passing through point A and (perpendicular to line) joining points P and Q is given

24.

Slope of line passing through points (2, 5) and (–3, 6) is

$m_1=\frac{-1}{m_1}=h$

$\frac{6-5}{-3-2}=\frac{1}{-5}$

So, slope of line perpendicular to line through (2, 5) and (–3, 6) is

$m_2=-\frac{1}{\left(\frac{-1}{5}\right)}=5$

$\therefore$ Equation of line with slope 5 and passing through (–3, 5) is

$\Rightarrow 5=\frac{y-5}{x-\left(-3\right)}=\frac{y-5}{x+3}$

$\Rightarrow 5\left(x+3\right)=y-5$

$\Rightarrow 5x+15-y+5=0$

$\Rightarrow 5x-y+20=0$

23.

Let RM be the median draw from vertex R so that M is the mid-point of line segment PQ. Then,

$=\left(\frac{2-2}{2},\frac{1+3}{2}\right)=\left(\frac{0}{2},\frac{4}{2}\right)$

= (0, 2)

So equation of line passing through R (4, 5) and M (0, 2) is

$y-5=\frac{2-5}{0-4}\left(x-4\right)$

$\Rightarrow y-5=\frac{-3}{-4}\left(x-4\right)$

$\Rightarrow 4\left(y-5\right)=3\left(x-4\right)$

$\Rightarrow 4y-20=3x-12$

$\Rightarrow 3x-4y+20-12=0$

$\Rightarrow 3x-4y+8=0$

23.

Given P = 5 and W = 30°

Using normal form, equation of line is

x cos W + y sin W = P