Class 11th

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

$\frac{x}{6}+\frac{y}{3}=1$

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

$\Rightarrow x+2y=6$

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

Hence, equation of line is 2x + y – 6 or x + 2y – 6 = 0

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

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

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

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

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 by

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

$\Rightarrow -\frac{1}{3}=\frac{y-\left(\frac{3}{1+n}\right)}{x-\left(\frac{2+n}{1+n}\right)}$

$\Rightarrow -\frac{1}{3}=\frac{y\left(n+1\right)-3}{x\left(n+1\right)-\left(2+n\right)}$

$\Rightarrow -\left[x\left(n+1\right)-\left(2+n\right)\right]=3\left[y\left(n+1\right)-3\right]$

$\Rightarrow -x\left(n+1\right)+\left(2+n\right)=3y\left(n+1\right)-3*3$

$\Rightarrow x\left(n+1\right)+3\left(n+1\right)y-\left(2+n\right)-9=0$

$\Rightarrow x\left(n+1\right)+3\left(n+1\right)y-n-2-9=0$

$\Rightarrow x\left(n+1\right)+3\left(n+1\right)y-n-11=0$

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

21

.Here (x₁, y₁) = (–1, 1) and (x₂, y₂) = (2, –4)

Using two point form, equation of line is

$y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)$

$\Rightarrow y-1=\frac{-4-1}{2-\left(-1\right)}\left(x-\left(-1\right)\right)$

$\Rightarrow y-1=\frac{-5}{2+1}\left(x+1\right)$

$\Rightarrow 3\left(y-1\right)=-5\left(x+1\right)$

$\Rightarrow 3y-3=-5x-5$

$\Rightarrow 5x+3y-3+5=0$

$\Rightarrow 5x+3y+2=0$

20.

The slope of the line is m = tan 30° = $\frac{1}{\mathrm{\surd 3}}$

And x-intercept, c = 2

Using slope intercept form, equation of line is

19.

Given m = –2

x-intercept, d = –3

Using slope intercept form, equation of line is

y = m (x – d)

$\Rightarrow y=\left(-2\right)\left[x-\left(-3\right)\right]$

$\Rightarrow y=\left(-2\right)\left(x+3\right)$

$\Rightarrow y=-2x-6$

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