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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 by

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

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

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

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

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

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

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

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

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)

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

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

$\Rightarrow y=-2x-6$

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

18.

The slope of the line is,  m = tan 75° – tan (45° + 30°)

17.

Given slope = m and $\left(x_0,y_0\right)=\left(0,0\right)$

By point slope from, equation of line is

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

$\Rightarrow m=\frac{y-0}{x-0}$

$\Rightarrow m=\frac{y}{x}$

$\Rightarrow mx=y$

$\Rightarrow$ $y=mx$

2.

Given,

$m=\frac{1}{2},\left(x_0,y_0\right)=\left(-4,3\right)$

By point slope form equation of line is

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

$\Rightarrow \frac{1}{2}=\frac{\left(y-3\right)}{x-\left(-4\right)}$

$\Rightarrow x+4=2\left(y-3\right)$

$\Rightarrow x+4=2y-6$

$\Rightarrow x-2y+4+6=0$

$\Rightarrow x-2y+10=0$

Exercise 9.2

15. 

For any points on x-axis the y-co-ordinate or the ordinate is always 0. Hence, the x-axis have the equation y = 0.

Similarly for xy points on y-axis the x-co-ordinate or the abscissa is always 0. Hence, the y-axis have the equation x = 0.

For B.Sc. Nursing at RIMT University, Gobindgarh, the semester fee is ?60,000, plus one-time charges: application fee ?1,000, registration fee ?15,000, and a refundable security deposit of ?8,000.

14.

From figure,

Slope of AB = $\frac{97-92}{1995-1985}$

$=\frac{5}{10}$

$=\frac{1}{2}$

As, A, B and C lie on same line

Slope of AB = Slope of BC

$\Rightarrow \frac{1}{2}=\frac{a-97}{2010-1995}=\frac{a-97}{15}$

$\Rightarrow 15=2\left(a-97\right)$

$\Rightarrow 15=2a-194$

$\Rightarrow 194+15=2a$

$\Rightarrow 209=2a$

$\Rightarrow a=\frac{209}{2}=104.5$

Hence, population in 2010 will be 104.5 crores.

13.

Since the three points say P (h, o), Q (a, b) and R (o, k) lie on a line.

Slope of PQ = slope of QR

$\Rightarrow \frac{b-o}{a-h}=\frac{k-b}{o-a}$

$\Rightarrow \frac{b}{a-h}=\frac{k-b}{-a}$

$\Rightarrow -ab=\left(k-b\right)\left(a-h\right)$

$\Rightarrow -ab=ka-kh-ab+bh$

$\Rightarrow ka+bh=kh$

Dividing both sides by kh we get,

$\Rightarrow \frac{ka}{kh}+\frac{bh}{kh}=\frac{kh}{kh}$

$\Rightarrow \frac{a}{h}+\frac{b}{k}=1$